I.-^ Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/advancedalgebraOOhawkrich ADVANCED ALGEBRA BY HEEBEET E. HAWKES, Ph.D. Assistant Professor of Mathematics in Yale University X GINN & COMPANY BOSTON . NEW YORK • CHICAGO • LONDON QA\54 e. Copyright, 1905, by H. E. HAWKES ALL BIGHTS KESERVED 613.1 GINN & COMPANY • PRO- PRIETORS . BOSTON . U.S.A. PEEFACE This book is designed for use in secondary schools and in short college courses. It aims to present in concise but clear form the portions of algebra that are required for entrance to the most exacting colleges and teclinical schools. The chapters on algebra to quadratics are intended for a review of the subject, and contain many points of view that should be presented to a student after he has taken a first course on those topics. Throughout the book the attention is concentrated on subjects that are most vital, pedagogically and practically, while topics that demand a knowledge of the calculus for their complete comprehension (as multiple roots, and Sturm's theorem) or are more closely related to other por- tions of mathematics (as theory of numbers, and series) have been omitted. The chapter on graphical r epresentation, has been intro- duced early, in the belief that the illumination which it affords greatly enlivens the entire presentation of algebra. The dis- cussion of the relation between , pairs of linear ftguntioT^ ^ g and pairs of straight lines is particularly su g gestive. _ In each chapter the discussion is directed toward a definite result. The chapter on theory of equations aims to give a simple and clear treatment of the method of obtaining the real roots of an equation and the theorems that lead to that iii 383513 vi CONTENTS CHAPTER n FACTORING SBOnOK ^ PAGE 28. Statement of the Problem 16 29. Monomial Factors 16 30. Factoring by grouping Terms 17 31. Factors of a Quadratic Trinomial 18 32. Factoring the Difference of Squares 20 33. Reduction to the Difference of Squares 20 34. Replacing a Parenthesis by a Letter 21 35. Factoring Binomials of the Form a" ± 6« 22 36. Highest Common Factor 22 S'7. H.C.F. of Two Polynomials 23 i^.:)i}adid"s Method of finding the H.C.F 23 39. Method of finding the H.C.F. of Two Polynomials ... 24 40. Least y^o>nmon Multiple 26 41. Second Rule for finding the Least Common Multiple ... 26 CHAPTER III FRACTIONS 42. General Principles . 27 43. Principle I 27 44. Principle II 27 45. Principle III 27 46. Reduction. . 27 47. Least Common Denominators of Several Fractions ... 28 48. Addition of Fractions 29 49. Subtraction of Fractions 29 60. Multiplication of Fractions 29 51. Division of Fractions 29 CHAPTER IV EQUATIONS 52. Introduction . 32 63. Identities and Equations of Condition 32 64. Linear Equations in One Variable 33 65. Solution of Problems . . 37 66. Linear Equations in Two Variables 40 57. Solution of a Pair of Equations 40 CONTENTS vu SECTION 68. Independent Equations 69. Solution of a Pair of Simultaneous Linear Equations 60. Incompatible Equations 61. R^um6 62. Solution of Problems involving Two Unknowns 63. Solution of Linear Equations in Several Variables PAGE . 41 • 42 . 42 43 • 45 47 CHAPTER V RATIO AND PROPORTION 64. Ratio 49 65. Proportion . 49 66. Theorems concerning Proportion 49 67. Theorem v 50 68. Mean Proportion 60 CHAPTER VI IRRATIONAL NUMBERS AND RADICALS 69. Existence of Irrational Numbers . 70. The Practical Necessity for Irrational Numbers 71. Extraction of Square Root of Polynomials . 72. Extraction of Square Root of Numbers . 73. Approximation of Irrational Numbers . 74. Sequences 75. Operations on Irrational Numbere 76. Notation 77. Other Irrational Numbers .... 78. Reduction of a Radical to its Simplest Form 79. Addition and Subtraction of Radicals . 80. Multiplication and Division of Radicals . 81. Rationalization 82. Solution of Equations involving Radicals 62 53 53 54 55 56 56 57 57 68 69 60 61 63 CHAPTER VII THEORY OF INDICES 83. Negative Exponents 66 84. Fractional Exponents 66 85. Further Assumptions 67 86. Theorem 67 87. Operations with Radical Polynomials 69 viii CONTENTS QUADRATICS AND BEYONI? i CHAPTER VIII QUADRATIC EQUATIONS SECTION PAGE 88. Definition 70 89. Solution of Quadratic Equations 70 90. Pure Quadratics 72 91. Solution of Quadratic Equations by Factoring .... 75 92. Solution of an Equation by Factoring 75 93. Quadratic Form 77 94. Problems solvable by Quadratic Equations 79 95. Theorems regarding Quadratic Equations 82 96. Theorem 83 97. Theorem 84 98. Nature of the Roots of a Quadratic Equation 84 CHAPTER IX GRAPHICAL REPRESENTATION 99. Representation of Points on a Line 87 100. Cartesian Coordinates 88 101. The Graph of an Equation 90 102. Restriction to Coordinates 91 103. Plotting Equations 91 104. Plotting Equations after Solution 93 105. Graph of the Linear Equation . "94 106. Method of plotting a Line from its Equation .... 96 107. Solution of Linear Equations, and the Intersection of their Graphs 97 108. Graphs of Dependent Equations 99 109. Incompatible Equations 09 110. Graph of the Quadratic Equation 100 111. Form of the Graph of a Quadratic Equation .... 101 112. The Special Quadratic ox^ + to = 103 113. The Special Quadratic aa;^ + c = 104 114. Degeneration of the Quadratic Equation 104 115. Sum and Difference of Roots 106 116. Variation in Sign of a Quadratic 107 CONTENTS ix CHAPTER X SIMULTANEOUS QUADRATIC EQUATIONS IN TWO VARIABLES SECTION PAGE 117. Solution of Simultaneous Quadratics Ill 118. Solution by Substitution Ill 119. Number of Solutions 113 120. Solution when neither Equation is Linear 114 121. Equivalence of Pairs of Equations 120 122. Incompatible Equations 121 123. Graphical Representation of Simultaneous Quadratic Equations 122 124. Graphical Meaning of Homogeneous Equations . . . 123 CHAPTER XI MATHEMATICAL INDUCTION 126. General Statement 125 CHAPTER XII BINOMIAL THEOREM 126. Statement of the Binomial Theorem 128 127. Proof of the Binomial Theorem 129 128. General Term 129 CHAPTER XIII ARITHMETICAL PROGRESSION 129. Definitions : .... 133 130. The nth Term 133 131. The Sum of the Series 134 132. Arithmetical Means 134 CHAPTER XIV GEOMETRICAL PROGRESSION 133. Definitions 137 134. The nth Term 137 135. The Sum of the Series 138 136. Geometrical Means 138 137. Infinite Series 140 CONTENTS ADVANCED ALGEBEA CHAPTER XV PERMUTATIONS AND COMBINATIONS SECTION PAGE 138. Introduction 143 139. Permutations 144 140. Combinations 146 141. Circular Permutations 149 142. Theorem 160 CHAPTER XVI COMPLEX NUMBERS 143. The Imaginary Unit 152 144. Addition and Subtraction of Imaginary Numbers . . . 163" 145. Multiplication and Division of Imaginaries 154 146. Complex Numbers 156 147. Graphical Representation of Complex Numbers .... 155 148. Equality of Complex Numbers . 155 149. Addition and Subtraction 156 150. Graphical Representation of Addition 156 151. Multiplication of Complex Numbers 157 152. Conjugate Complex Numbers . . . . . . . 158 153. Division of Complex Numbers 158 154. Polar Representation 160 155. Multiplication in Polar Form 160 156. Powers of Numbers in Polar Form 161 157. Division in Polar Form 162 158. Roots of Complex Numbers 162 CHAPTER XVII THEORY OF EQUATIONS 159. Equation of the nth Degree 166 160. Remainder Theorem 166 161. Synthetic Division . . 167 162. Proof of the Rule for Synthetic Division 169 163. Plotting of Equations . . 170 164. Extent of the Table of Values 171 ^ CONTENTS xi SIECTION PAGE fl65. Roots of an Equation 172 . Number of Roots 172 167. Graphical Interpretation 174 ^ 168. Imaginary Roots . . . 174 . Graphical Interpretation of Imaginary Roots .... 175 170. Relation between Roots and Coefficients 177 171. The General Term in the Binomial Expansion .... 178 172. Solution by Trial . 178 173. Properties of Binomial Surds 179 174. Formation of Equations . . 180 176. To multiply the Roots by a Constant 183 176. Descartes' Rule of Signs 186 177. Negative Roots 189 178. Integral Roots 190 179. Rational Roots 190 180. Diminishing the Roots of an Equation . . ... . 191 181. Graphical Interpretation of Decreasing Roots .... 193 182. Location Principle 194 183. Approximate Calculation of Roots by Horner's Method . . 195 184. Roots nearly Equal 200 CHAPTER XVIII DETERMINANTS 185. Solution of Two Linear Equations 203 186. Solution of Three Linear Equations 204 187. Inversion 208 188. Development of the Determinant 208 189. Number of Terms 210 190. Development by Minors 210 191. Multiplication by a Constant 213 192. Interchange of Rows and Columns 213 193. Interchange of Rows or Columns 214 194. Identical Rows or Columns . . . . . . . 215 195. Proof for Development by Minors 215 196. Sum of Determinants 216 197. Vanishing of a Determinant 217 198. Evaluation by Factoring 218 199. Practical Directions for evaluating Determinants . ... 219 200. Solution of Linear Equations 221 ^ 201. Solution of Homogeneous Linear Equations .... 223 4^^ — -• xii CONTENTS CHAPTER XIX PARTIAL FRACTIONS SECTION .^ PAGE 202. Introduction 226 203. Development when (f){x)= has no Multiple Roots . . . 225 204. Development when (p{x)=0 has Imaginary Roots . . . 229 205. Development when <p (x) = {x — ar)» 232 206. General Case 233 CHAPTER XX LOGARITHMS 207. Generalized Powers 235 208. Logarithms . ' 236 209. Operations on Logarithms 237 210. Common System of Logarithms 239 211. Use of Tables 241 212. Interpolation 242 213. Antilogarithms 243 214. Cologarithms 246 215. Change of Base 247 216. Exponential Equations 261 217. Compound Interest 263 CHAPTER XXI CONTINUED FRACTIONS 218. Definitions . ' 256 219. Terminating Continued Fractions 256 220. Convergents 258 221. Recurring Continued Fractions 260 222. Expression of a Surd as a Recurring Continued Fraction . 263 223. Properties of Convergents 266 224. Limit of Error 267 CHAPTER XXII INEQUALITIES 225. General Theorems . .269 226. Conditional Linear Inequalities 271 227. Conditional Quadratic Inequalities 271 CONTENTS xiii CHAPTER XXm VARIATION SECTION ~ PAOB 228. General Principles . .273 CHAPTER XXIV PROBABILITY 229. Illustration 276 230. General Statement 276 CHAPTER XXV SCALES OF NOTATION 231. General Statement 279 232. Fundamental Operations ... .... 280 233. Change of Scale ..... .... 281 234. Fractions 282 236. Duodecimals . , 284 ik ADVANCED ALGEBRA ALGEBRA TO QUADRATICS CHAPTER I FUNDAMENTAL OPERATIONS 1. It is assumed that the elementary operations and the mean- ing of the -usual symbols of algebra are familiar and do not demand detailed treatment. In the following brief exposition of the formal laws of algebra most of the proofs are omitted. 2. Addition. The process of adding two positive integers a and h consists in finding a number x such that a -\-h = x. For any two given positive integers a single sum x exists which is itself a positive integer. 3. Subtraction. The process of subtracting the positive num- ber h from the positive number a consists in finding a number x such that h -\- x = a. (1) This number x is called the difference between a and h and is denoted as follows : a — 6 = a;, a being called the minuend and h the subtrahend. If a > ft and both are positive integers, then a single posi- tive integer x exists which satisfies the condition expressed by equation (1) 1 . AL'G^BE a: to * ^JJ ADR ATI€S ^ li a <h, then x is not a positive integer. In order that the pro- cess of subtraction may be possible in this case also, we introduce negative numbers which we symbolize by (—a ), (— b), etc. When in the difference a~b,ais less than b, we define a — b = (— (b — a)). The processes of addition and subtraction for the negative numbers are defined as follows : a -^ (— b) = a — b. a — (— b) = a -^ b. 4. Zero. If in equation (1), a = b, there is no positive or nega- tive number which satisfies the equation. In order that in this case also the equation may have a number satisfying it, we intro- duce the number zero which is symbolized by and defined by the equation a -\- = a, or a — a = 0. The processes of addition and subtraction for this new number zero are defined as follows, where a stands for either a positive or a negative number 0-a = -a. ± = 0. 5. Multiplication. The process of multiplying a by i consists in finding a number x which satisfies the equation a-b = X. * The symbol for a positive integer might be written (+ a), (+ 6), etc., consistently with the notation for negative numbers. Since, however, no ambiguity results, we omit the + sign. Since the laws of combining the + and — signs given in this and the folloM'ing paragraphs remove the necessity for the parentheses in the notation for the negative number, we shall omit them where no ambiguity results. FUNDAMENTAL OPERATIONS 3 When a and h are positive integers cc is a positive integer which, may be found by adding a to itseK h times. When the numbers to be multiplied are negative we have the following laws, (— a) • (— 5) = a • 5, Oa = «;.0 = 0, (1) where «; is a positive or negative number or zero. These symbolical statements include the statement of the following Principle. A 'product of numbers is zero when and only when one or more of the factors are zero. This most important fact, which we shall use continually, assures us that when we have a product of several numbers as a-b-G- d =^ Bj first, if e equals zero, it is certain that one or more of the num- bers a, bj c, or d are zero ; second, if one or more of the numbers a, b, c, or d are zero, then e is also zero. 6. Division. The process of dividing ahj p consists in finding a number x which satisfies the equation X-p = a, . (1) where a and jS are positive or negative integers, or a is 0. When a occurs in the sequence of numbers • ••-.?A -2A -Ap, ft-2A 3A •••, a; is a definite integer or 0, that is, it is a number such as we have previously considered. If a is not found in this series, but is between two numbers of the series, then in order that in this case the process may also be possible we introduce the fraction which we symbolize by a -*- /8 or - and which is defined by the equation 4 ALGEBRA TO QUADRATICS The operations for addition, subtraction, multiplication, and divi- sion of fractions are defined as follows : fi 8 Py W Further properties of fractions are the following : a 1 — = — > where S is any number, (5) j8 SP The last two equations are expressed verbally as follows : Both numerator and denominator of a fraction may be multi- plied by any number without changing the value of the fraction. Changing the sign of either, numerator or denominator of a fraction is equivalent to changing the sign of the fraction. The laws of signs in multiplication given on p. 3 may now be assumed to hold when the letters represent fractions as well as integers. j- ,.-, . Thus for example ~ I t ) ' "~ ac M The positive or negative number a may be written in the fractional form a 1* 7. Division by zero. If in equation (1), ^Q>^ p=i 0, there is no single number x which satisfies the equation, since by (1), § 5, whatever value x might have, its product with zero would be zero. FUNDAMENTAL OPERATIONS 5 Thus division by zero is entirely excluded from algebraic pro- cesses. Before a division can safely be performed one must be assured that the divisor cannot vanish. In the equation 4 = 2.0, if we should allow division of both sides of our equation by zero, we should be led to the absurd result 4 = 2. 8. Fundamental operations. The operations of addition, sub- traction, multiplication, and division we call the four fundamental operations. Any numbers that can be derived from the number 1 by means of the four fundamental operations we call rational num- bers. Such numbers comprise all positive and negative integers and such fractions as have integers for numerator and denominator. Positive or negative integers are called integral numbers. 9. Practical demaild for negative and fractional numbers. In the preceding discussion negative numbers and fractions have been introduced on account of the mathematical necessity for them. They were needed to make the four fundamental opera- tions always possible. That this mathematical necessity corre- sponds to a practiqal necessity appears as soon as we attempt to apply our four operations to practical affairs. Thus if on a certain day the temperature is + 20° and the next day the mer- cury falls 25°, in order to express the second temperature we must subtract 2^ from 20. If we had not introduced negative numbers, this would be impossible and our mathematics would be inapplicable to this and countless other everyday occurrences. 10. Laws of operation. All the numbers which we use in algebra are subject to the following laws. ' .Commutative law of addition. This law asserts that the value of the sum of two numbers does not depend on the order of summation. Symbolically expressed, a -\- h = h -\- aj where a and h represent any numbers such as we have presented or shall hereafter introduce. 6 ALGEBRA TO QUADRATICS Associative law of addition. This law asserts that the sum of three numbers does not depend on the way in which the numbers are grouped in performing the process of addition. Symbolically expressed, a 4- (^ + c) = (a + &) + c = a 4- * + c. Commutative law of multiplication. This law asserts that the value of the product of two numbers does not depend on the order of multiplication. Symbolically expressed, a-b = b-a. Associative law of multiplication. This law asserts that the value of the product of three numbers does not depend on the way in which the numbers are grouped in the process of multiplication. Symbolically expressed, (^a-b) -c = a-(b'C)= a-b-c. Distributive law. This law asserts that the product of a single number and the sum of two numbers is identical with the sum of the products of the first number and the other two numbers taken singly. Symbolically expressed, a- (b -\- c)= a-b -{- a-c. All the above laws are readily seen to hold when more than three numbers are involved. 11. Integral and rational expressions. A polynomial is inte- gral when it may be expressed by a succession of literal terms, no one of which contains any letter in the denominator. Thus 4 a;5 _ a;3 _ 2 x2 _ J x + 1 is integral. The quotient of two integral expressions is called rational. ^^ a;2_2a; + 3. ^. , Thus z — 18 rational. a — 7 12. Operations on polynomials. We assume that the same formal laws for the four fundamental operations enunciated in § § 2-6 and the laws given in § 10 hold whether the letters in the symbolic statements represent numbers or polynomials. FUNDAMENTAL OPERATIONS 7 In fact the literal expressions which we use are in essence nothing else than numerical expressions, since the letters are merely symbols for numbers. When the letters are replaced by numbers, the literal expressions reduce to numerical expressions for which the previous laws have been explicitly given. 13. Addition of polynomials. For performing this operation we have the following EuLE. Write the terms with the same literal part in a column. Find the algebraic sum of the terms in each column, and write the results in succession with their proper signs. When the polynomials reduce to monomials the same rule is to be observed. EXERCISES Add the following : 1. 3a262_2a6 + 6a26-a; 4. ab - 2 a^h'^ - 11 a'^b + 9 a ; 2a^b -ab -2a; Sam- 4:db -6a. Solution: ZaW-2ab+ 6a^b- a -2a262 + 4a6- lla26 + 9a - a6 + 2 a26 - 2 a Sa^-iab -6a 9am -Sab- Sa^b 2. 21a-246-8c2; 16c + 17 6 + 6c2 - 20a ; 186 -18c. 3. x* -6a;2- 8x-l; 2x3 + 1; 6x'^ + 1 x + 2 ; x^ - x^ + x - 1. 4. 9(a + 6)-6(6 + c) + 7(a + c); 4(6 + c) - 7(a + 6) - 8(a + c) ,• (a + c) - (a + 6) + (6 + c). 5. a2-4a6 + 62 + a + 6-2; 2a2 + 4a6- 362- 2a -26 + 4; 3a2- 5a6-462 + 3a + 36 - 2; 6a2 + 10a6 + 562 + a + 6. 14. Subtraction of polynomials. For performing this opera- tion we have the following EuLE. Write the subtrahend under the minuend so that terms with the same literal part are in the same column. To each term of the minuend add the corresponding term of the subtrahend, the sign of the latter having been changed. It is generally preferable to imagine the signs of the subtrahend changed rather than actually to write it with the changed signs. 8 ALGEBRA TO QUADRATICS EXERCISES 1. From a262 - 3 a^ft + 8 a6 + 6 6 subtract 9 a'^h'^ - 6 a6 + 4 a^ft + a. Solution : a%'^ - 3 a26 + 8 a& + 6 6 -9a262:f 4a26- 6a6 +a - 8a262 ^ 7 a26 + 14a6 + 66 - a 2. From 6 ahx — 4 win + 5 x subtract 3 mn + 6 ax — 4 a6x. 3. From m -{■ an + bq subtract the sum of cm + dn + {b — a)q and {a — b)q — {a -\- d)n — cm. 4. From the sum of | a + y^ 6 + | c and — 6 — c — a subtract ^6 — |c + ^a. 5. From the sum of 2 x^ — 3 x + 4 and x* — f x — i subtract x^ — | x^ -3|x + 3i. 15. Parentheses. When it is desirable to consider as a single symbol an expression involving several numbers or symbols for numbers, the expression is inclosed in a parenthesis. This paren- thesis may then be used in operations as if it were a single number or symbol, as in fact it is, excepting that the operations inside the parenthesis may not yet have been carried out. EuLE. When a single parenthesis is preceded hy a -\- sign the parenthesis may he removed, the various terms retaining the same sign. When a single parenthesis is preceded hy a — sign the parent- thesis may be removed, providing we change the signs of all the terms inside the parenthesis. When several parentheses occur in an expression we have the following EuLE. Remove the innermost parenthesis, changing the signs of the terms inside if the sign preceding it is minus. Simplify, if possible, the expression inside the new inn^rwjost parenthesis. Repeat the process until all the parentheses are removed. It is in general unwise to shorten the process by carrying out some of the steps in one's head. The liability to error in such attempts more than offsets the gain in time. FUNDAMENTAL OPERATIONS EXERCISES Remove parentheses from the following : 1. 6 _ pa -[26 + (4a -2a -6) -66]}. Solution: 6 - {9a - [26 + (4a - 2a - 6) - 66]} = 6- {9a -[26 + (4a -2a + 6) -66]} = 6-{9a-[2 6+ (2a + 6)-6 6]} = 6 _ [9 a - (2 6 + 2 a + 6 - 6 6)] = 6- [9a -(2a -36)] = 6 -(9a -2a + 36) = 6 - (7 a + 3 6) = 6-7a-36 = - 7 a - 2 6. 2- -{-[-[-(-(-I))]]}- 3. a2 + 4 - {6 - [- (a2 - 6) + 1]}. 4. i{i-|[f-ia-i-f-T\) + |]-^}- 5. x2 - {2/2 _ [4x + 3(y - 9x(y - x)) + 9y (x - y)]}. 6. Find the value of a - {5 6 - [a - (3 c - 3 6) + 2 c - 3 (a - 26 -c)]} i^ when a = — 3, 6 = 4, c = — 5. 16. Multiplication. It is customary to write a - a = a^ ; aaa = a^] a - a • • a = a"*. We have tlien by tlie associative n terms law of multiplication, § 10, a^ - a^ = (a • a) (a - a ' a) = a a • a ' a • a = a^, or, in general. where m and n are positive integers. Furthermore, (I) (a')" = a"-a'---a' = a'-"'. (II) m terms Finally, a" • J" = (a • bf. (Ill) The distinction between (a**)"* and a""* should be noted carefully. Thus (28)* = 82 = 64, while 28* =2^ = 512. Equation (I) asserts that the exponent in the product of two powers of any expression is the sum of the exponents of the factors. Hence we may multiply monomials as follows : 10 ALGEBRA TO QUADRATICS I Rule. Write the product of the numerical coefficients, followed , by all the letters that occur in the multiplier and multiplicand, each having as its exponent the sum of the exponents of that letter in the multiplier and multiplicand. Example. 4 a2^i0c# • (- 16 a'^bd'^) = - 64 a^b^^cd^\ 17. Multiplication of monomials by polynomials. By the distributive law, § 10, we can immediately formulate the EuLE. Multiply each term of the polynomial by the monomial and write the resulting terms in succession. Example. 9 a'^h'^ - 2 a& + 4 a62 - a + &* 3a26 27 a463 _ 6 aW + 12 a^h^ - 3 a^fe + 3 aW * 18. Multiplication of polynomials. If in the expression for the distributive law, a(c + d)= ac •{- ad, we replace a by a + 5, we have (a -\- h) (c -\- d) = ac -\- be -\- ad -{- bd, which affords the EuLE. Multiply the mult^icand by each term of the multi- plier in turn, and write the partial products in succession. To test the accuracy of the result assume some convenient numer- ical value for each letter, and find the corresponding numerical value of multiplier, multiplicand, and product. The latter should be the product of the two former. EXERCISES 1. Multiply and check the following: (a) 2 a2 + a6 + 4 62 + 5 and a-h-\- ah. Check : Solution : Let a = 6 = 1 2a2+ a6+462 + 6 = 8 a - h + a6 = 1 2a8+ a26 + 4a&2 + a6 -2a26- a62 -463_ -62 + a62 + 2a86 + a262 + 4a68 2a8 - a26 + 4a62 + a6 - 46^ - 62 -f- 2a86 + a262 + 4a68 = 8 FUNDAMENTAL OPERATIONS H (b) 6 abx^ and 4 a'^b^x. (c) ^^and -OxV^. o (d) 3 a62x - ^ 6x4 and 6 a^cx. (e) x2« + ?/26 ^ x^ and x« — y^ (f ) a^ + ab + l^ and a^ + ac + c2. (g) x^y^'^j a;»-3yin4-4^ and ic^y2m-2^ (h) xP-3 4- xP-2 + 1 and a;3 _ a;2 _ 1, (i) 8 a26c, - a62, _ 7 62, - — a^c*, and -. w ' 4 ' ' 14 ' 6 (j) ax* — 2 a2x3 — X + 4 a and — x + 2 a. (k) x« + * + x2« + x2& 4- a;3o-& and x«-^ — 1. (1) 15x* - 11x3 + 6x2 + 2x - 1 and - 3x2 - 1. (m) 4 x* — 8 xy^ + | x2y2 _ 3 x^y — x — y and — 42 xy. 2. Expand (x + y)*. 3. Expand and simplify (a;2 + 2/2 + 22)2 _ (X + y + 2r) (X + y - z) (x + 2 - y) (y + 2 - X). 19. Types of multiplication. The following types of multi- plication should be so familiar as merely to require inspection of the factors in order to write the product. EuLE. The product of the sum and difference of two terms is equal to the square of the terms with like signs minus the square of the terms which have unlike signs. Examples. (a - 6) (a + 6) = a?^ — 62. (4x2 - 3y2)(4x2 + 3y2) = 16x4 - 9y4. 20. The square of a binomial. This process is performed as follows : KuLE. The square of a binomial, or expression in two terms, is equal to the sum of the squares of the two terms plus twice their product. Examples. (x + y)2 = x2 + y2 + 2 xy. (2a-36)2=:4a2 + 962- 12a6. 12 ALGEBRA TO QUADRATICS 21. The square of a polynomial. This process is performe! as follows : Rule. The square of any polynomial is equal to the sum of the squares of the terms plus twice the product of each term hy each term that follows it in the polynomial. Example, (a + 6 + c)2 = a^ + 62 + c^ + 2 a6 + 2 ac + 2 6c. 22. The cube of a binomial. This process is performed as follows : Rule. The cube of any binomial is given hy the folloimng expression : ^^ _^ ^y ^ ^s _^ ^ ^2j -\.3ab^ + h\ EXERCISES Perform the following processes by inspection. 1. (a-6 + c)2. 2. (a4-66)2. 3. (2x'-i-l)2 4. (a2-62)3. 5. (2x'-i-l)3. 6. (l-8a;22/)2. 7. (X2-2X + 1)2. 8. (a;2 - y2 + ^2)2. 9. (2a-26-c)2. 10. (x8-2x-l)2. 11. (ai'-3-6p + 3)2. 12. (-6x2y + 4xy2)8. 13. {xp - 2/9) {xp + y^). 14. (- 6x22/ + 4 xy^)^. 15. (3x+ 2 2/)(3x-2 2/). 16. (- 3ax2 + 2ax -6)2. 17. {-3x^y-^lz^){Sx'^ + ^z^). 18. (_ 4 - 6a26) (- 4 + 6a26). 19. (2 a -2^. 20. (2 a-?)'. 23. Division. By the definition of division in § 6, we have a = —} a'^ = — ) a^ = —z'y . a a a'' or, in general, ^ a"-™ = — -> a"" where n and m are positive integers and n> m. If n = m, we preserve the same principle and write a«-» = a« = — = 1, (1) FUNDAMENTAL OPERATIONS 13 24. Division of monomials. Keeping in mind the rule of signs for division given in § 6, we have the following EuLE. Divide the numerical coefficient of the dividend hy that of the divisor for the numerical coefficient of the quotient^ keeping in mind the rule of signs for division. Write the literal part of the dividend over that of the divisor in the form of a fraction^ and perform on each pair of letters occurring in both numerator and denominator the process of division as defined hy equation (I) in the preceding paragraph. Example. Divide 12 a'^lA^cH by - 6 a^hc^d^. 12 a^6"c2d _ 2 bio 25. Division of a polynomial by a monomial. This process is performed as follows : Rule. Divide each term of the polynomial hy the monomial and write the partial quotients in succession. Example. Divide 8 a^lP - 12 a^h^ by 2 aW. 8a266 _ 12 gcftg _ 463 _ 6a» 2a363 2a363 ~ a & * 26. Division of a polynomial by a polynomial. This process is performed as follows : Rule. Arrange hoth dividend and divisor in descending powers of some common letter {called the letter of arrangement)- Divide the first term of the dividend hy the first term of the divisor for the first term of the quotient. Multiply the divisor hy this first term of the quotient and subtract the product from the dividend. Divide the first term of this remainder hy the first term of the divisor for the second term of the quotient, and proceed as before until the remainder vanishes or is of lower degree in the letter of arrangement than the divisor. 14 ALGEBRA TO QUADRATICS When the last remainder vanishes the dividend is exactly divisible by divisor. This fact may be expressed as follows: dividend _ . . . ,. . = quotient. divisor When the last remainder does not vanish we may express the result of division dividend .. , , remainder -^rr-. = quotient -\ -— : divisor divisor The coefficients in the quotient will be rational numbers if those in both divi- dend and divisor are rational. EXERCISES Divide and check the following : 1. 8a8 + 6a26 + 9a62 + 963by4a + &. Solution : 4a4-&| 8 a^ + 6 a26 + 9 a62 + 9 h^ \2 a^ + ab + 2lfl 8 a3 + 2 a'^b 4 a26 + 9 a62 4a26+ am 8a&2 + 9&3 8a&2 + 268 76» 7 63 . ■ Result : 2a^-\-ah-{-2,b^ + 1 4a + 6 • Check : Let a = h =1. Dividend = 32, divisor = 5, quotient = 6§. 32 - 5 = 6f . 2. xi2 - yi2 by x3 - 2/3. 3. 2 x2 - 6x + 2 by X - 2. 4. xi2 — 2/12 i3y aj4 _ 2/*. 5. x^ — 2/6 by x^ + xy -\- y^. 6. a8 - a2 + 2 bya + 1. 7. - 63x*y32;2 ^y _ gx^y^z. 8. .x2 - X - 30 by X + 5. 9. 4 a26 - 6 a62 _ 2 a by - 2a. • 10. 16 a264cii by - 2 a^¥c\ 11. i x2 - 3^ x - | by 1^ x + T»ff. 12. ax2 + (a2 - 6) X - a6 by X + a. 13. ax« + 6x»-i + cx«-2 - dx«-3 by x'. 14. 16 a2x22/2 _ 8 ax32/2 - 4 x*?/ by - f xy. 15. ISai'b^ -\- 6aP + 2?^+3_ 9 ap + ^ft^ by 3 apft*. 16. a2 - 2 a6 - 4 c2 + 8 6c - 3 62 by a - 2 c + 6. 17. x* — (d + 6 + c) x2 -f {ab + ac + 6c) x - a6c by x - a. 18. 2j/2 _ 6x2 + i/xy + V-x - -^y + 1 by 2x + |y - f. 19. x8 - 2 x22/ - x2 + 2/2x f 2 xy - y - 2/2 + 2 by X - y + 1. 20. 3x8 + 6x22/ + 9x2 + 2x2/2 + 5^3 + 22/ + 6y2 + 3 by x + 22/ + 3. FUNDAMENTAL OPERATIONS 15 27. Types of division. The following types of division, which may be verified by the rule just given for any particular integral value of rij should be so familiar that they may be performed by inspection. (a^n _ ^,2n) ^ (^^» -t ^n^ = «»» ^ ^«. (I) (a" + b^) -^ (a + 5) = a"-i - a^-% + a^-%'' + b^-\ (II) where n is odd. (a" -. h^) ^(a-b)^ a"-i 4- a"-'^ + a«-3^>2 + . . . + j«-i^ (III) where n is odd or even. EXERCISES Give by inspection the results of the following divisions. 1. a« - 1 by a -r l: 2. a^ + 1 by a + 1. 3. x' + 128 by ic + 2. 4. x^ + y^hy x-\- y. 6. x^ — y^ by x^ -{■ y^. 6. x^ — y* hj x — y. 7. x8 - ^8 by X* - y*. 8. a2m _ 1 by a - 1. 9. a2«+^ - 1 by a - 1. 10. 27a9 + 868 by 3^8 + 26. 11. 8x8-27 by 2a:- 3. 12. 4a2 - 25668 by 2a + 1664. 13. 16 a* - 256 by 4 a^ + 16. 14. 27 ai^ - 64 612 by 3 a* - 4 6*. CHAPTER II FACTORING 28. Statement of the problem. The operation of division con- sists in finding the quotient when the dividend and divisor are given. The product of the quotient and the divisor is the divi- dend, and the quotient and the divisor are the factors of the dividend. Thus the process of division consists in finding a second factor of a given expression when one factor is given. The process of factoring consists in finding all the factors of a polynomial when no one of them is given. This operation is in essence the reverse of the operation of multiplication. We shall be concerned only with those factors that have rational coefficients. 29. Monomial factors. By the distributive law, § 10, ah -{• ac = a(}) -\- c). This affords immediately the Rule. Write the largest monomial factor which occurs in every term outside a jpareiithesis which includes the algebraic sum of the remaining factors of the various terms. EXERCISES Factor the following : 1. 6 Q?Wc + 9 alPc^ - 15 a*6c7. Solution : 6 a'^hH -t- 9 ah^c^ - 15 a^hc^ = 3 a6c (2 at^ -J- 3 ftScS - 5 a^<fi). 2. 14 anx — 21 6nic — 7 n. 3. 121 a'^hH - 22 a^hc'^ + H oJb'^cK 4. 5xV - 10x8?/8 - SxV - 15x2y2. 5. 21 ahn + 6 obH"^ - 18 a'^hv?- + 15 a'^h'^n. 6. 10 al^cmx - 5 ab^cy + 6 ab'^cz - 15 abc^m^. 7 a^xV - 49 ax^y* + 14 axy^z^ - 21 a^'^y^. 8. 45 a*62c8d - 9a6*c«d8 -I- 27 a^bc*d^ - 117 a^l^cd*. 16 FACTORING 17 30. Factoring by grouping terms. If in the expression for the distributive law, ac + bc = {a-\- b) c, we replace chj c -\- d, we have a(G -\- d) -\- h (c -\- d) = ac -\- ad -\- be -\- bd. We may then factor the right-hand member as follows : ac_ -^ ad -\- be -\- bd = a (c -\- d) -\- b {c -\- d) = {a -{- b) (c -\- d). This affords the EuLE. Factor out any monomial expression that is common to each term of the polynomial. Arrange the terms of the polynomial to be factored in groups of two or more terms each, such that in each group a monomial factor may be taken outside a parenthesis which in each case contains the same expression. Write the algebraic sum of the monomial factors that occur outside the various parentheses for one factor, and the expres- sion inside the parentheses for the other factor. EXERCISES Factor the following : 1. 4a3&-6a262_4a4 + 6a63. Solution : 4 a^b - 6 a'^h'^ - 4 a* + 6 aft* = 2 a (2 a26 - 3 a62 _ 2 a3 + 3 63) = 2 a (2 a26 - 2 a^ - 3 a62 + 3 &») = 2 a [2 a2 (6 ^a) + 3 62 (6 _ a)] = 2 a (6 - a) (2 a2 + 3 62). 2. x2-(3a + 46)x + 12a6. Solution : x2 - (3 a + 4 6) x + 12 a6 = x2 - 3 ax - 4 6x + 12 a6 = X (x — 3 a) — 4 6 (x — 3 a) = (x-46)(x-3a). 3. 2 ax - 3 6y - 2 6x + 3 ay. 4. 56 a2 - 40 a6 + 63 ac - 45 6c. 5. a''x2 — 6'?y3 _ f/jx^ + apy\ 6. 91 x2 - 112 mx + 65 nx - 80 mru 7. ax - bx -[■ ex + ay - by -\- cy. 8. 2 ax - 6x - c6 + 2 a6 + 2 ac - 62. 18 ALGEBRA TO QUADRATICS 9. 2x6 - 3x8 + 2x« - 3. 10. x^ + x^ + x + 1. 11. acx"^ - hex + adx-bd. 12. 2 6x - x^ - 4 & + 2 x. 13. 3x3 _ 12x82/2 _ 4y2 + 1. 14. 4x2 - (8 + b)x + 2 6. 15. x* - (4m + 9n)x2 + 36 mn. 16. x* - (2m + 3n)x2 + 6mn. 17. a2x-ac+a6y-a62x-63y +c62. 18. 18 a» - 2 ac*&6 _ g a26 + c^b^. 19. 2ax — 6ay + a-26x + 56y-6. 20. 2ax-ay-26x + 4cx— 2cy+62/. 21. 2 a2x« + 4 a2x* + 2 a2x2 + 4 a^. 22. 8x2 - 2 ox - 12 xz + 3az + 4xy - ay. 31. Factors of a quadratic trinomial. In this case we cannot factor by grouping terms immediately, as that method is inappli- cable to a polynomial of less than four terms. We observe, how- ever, that in the product of two binomial expressions. (mx + n) (px + qy= tnp^J -h {mq + w^)ic -|- ngy the coefficient of x is the sum of two expressions mq and np^ whose product is equal to the product of the coefficient of x^ and the last term, that is, mq • np = m,p • nq. Thus, to factor the right-hand member of this equation, we may remove the parenthesis from the term in x and use the principle rf grouping terms. Thus mpx^ -f (mq + np) q? + nq = TYipx^ -j- m.qx + npx -f- nq = mx (px -\- q)+ n (px -f q) = (mx + n) (px -\- q). This affords the Rule. Write the trinomial in order of descending powers of X (or the letter in which the expression is quadratic). Multiply the coefficient of a^ by the term not involving x, arid find two factors of this product whose algebraic sum is the coefficient of x. Replace the coefficient of x by this sum and factor by group- ing terms. FACTORING 19 Factoring a perfect square is evidently a special case under this method. Thus factor x2 + 6x + 9. 1-9=9. 3 + 3 = 6. x2 + (3 + 3)x + 9 = x2 + 3x + 3x + 9 = a;(x + 3) + 3(x + 3) = (X + §) (X + 3) = (X + 3)2. One will usually recognize when a trinomial is a perfect square, in which case the factors may be written down by inspection. + 2 = 8. EXERCISES Factor the following : ( 1. 3x2 + 8x + 4. / Solution : 3-4 = 12. / 6 H 3x2 + 8x + 4 = 3x2 + (6 + 2)x + 4 = 3x2 + 6x + 2x + 4 = 3x(x + 2) + 2(x + 2) = (3x + 2)(x + 2). 2. 8x2- 146X + 352. Solution : 8. 362:.. 2462. 8x2-146x + 362 26 -.126 = -146. = 8x2-(2 6 + 12 6)x + 362 = 8x2-126x-26x + 362 = 4x(2x-36)-6(2x-36) = (4x-6)(2x-36). 3. 28x2 -3x- 40. Solution : 28 • (- 40) = - 1120. The factors of 1120 must be factors of 28 and 40. We seek two factors of 1120, one of which exceeds the other by 3. We note that since 40 exceeds 28 by more than 3, one factor must be greater and the other less than 28 and 40 respectively. Since 4 • 7 = 28 and 5 ■ 8 = 40, we try 5 • 7= 35 and 4 • 8 = 82, which are the required factors of 1120. 28x2 -3x- 40 = 28x2-(35-32)x-40 = 28 x« - 35 X + 32 X - 40 = 7x(4x-6) + 8(4x-6) = (7x + 8)(4x-5). 20 ALGEBRA TO QUADRATICS 4. x2-6x + 9. 5. 2x2 + re -6. 6. 2x2 - X - 6. 7. 2x2 + X - 91. 8. x2 + X - 182. 9. 9x2 - 2x - 7. 10. 2x2 + 5a; + 3. 11. x2 - llx + 18. 12. 3x2-10x-8. 13. 6x2 + 17 x + 7. 14. 16x2 + 4x- 3. 15. 7?/2-4?/-ll. 16. a2-6a6 + 962. 17. 5x2-12x + 4. 18. 18x2-73x + 4. 19. 27x2 + 3x- 2. 20. 24x2 - 31x - 16. 21. 21x2 - 31x + 4. 22. 12 x2 + 60 X - 72. 23. x* - 3 az^-A 2 a^. 24. 9x2-18ax-7a2. 25. 10x2-63x-13. 26. x4y« - 12 x2?/3 + 36. 27. 4x2p-16xi'- 81. >v 28. 2 x8 - 17 6x2 + sif'^x. 29. 4 a2 + 12 a6 + 9 62. 30. 6 a2x2 - 2 a6x - 7 62. ' 3 1. 10 x* - 16 a2x3 - 100 x^a^. 32. 4 aa"* + 16 a^b» + 16 62«. 33. 4 a^x^y* - 20 a6xy2z + 25 62z2. 32. Factoring the difference of squares. Under the method of the preceding paragraph we may factor the difference of squares. Thus to factor x^ — b^ we observe that the product of the coefficient of x^ and the constant term is 1 . (_ b^) = _ b^. Since the coefficient of x. is zero, we have -b-\-b = 0. Hence (x -\- b)(x — b)— x^ — b\ EuLE. Extract the square root of each term. The sum of these square roots is one factor, and their differ- ence is the other. Example. Factor 9a2xV - 16 68c2. 9 a2x«2/* - 16 68c2 = (3 axV + 4 64c) (3 ax-3y2 _ 4 54c) . 33. Reduction to the difference of squares. The preceding method may be used when the expression to be factored becomes a perfect square by the addition of the square of some expression. FACTORING 21 EXERCISES Factor the following : 1.44 ^4 6*. Solution r a* + 4 6* = a^ + 4 0,262 + 4 54 _ 4 ^^252 = (a2 + 2b2)2_4a262 = (a2 + 2 62 _ 2 ab) {a^'-\- 2 62 + 2 db). 2. 1 - a*. - 3. a* + 4. 4. x^ — X. 5. x«y4 + 4 x2. 6. 4a;4+.2/*. 7. 4a2-2562. 8. x* + ic2 ^. 1. 9 ig ^254 _ 3.4. 10. x4 + 9x2 + 81. 11. 4a2p - 962c2«. 12. a2p + 3 - 16a36*. 13. ar*« + x2« + 1. 14. 36x2^4^8 - 49 w2ui6. 15. x4 - 13x2 + 36. 16. nty^ X IG mx* - lij m'x^y'^. 1 7. 9 x* + 8 x^^ + 4 y^. 34. Replacing a parenthesis by a letter. Any of the preceding methods may be applied when a polynomial appeare in place of a letter in the expres- sion to be factored. It is frequently desirable for simplicity to replace such a polynomial by a letter, and in the final result to restore the polynomial. EXERCISES Factor the following : 1. 2 ax2 - 2 6x2 - 6 ax + 6 6x - 8 a + 8 6. Solution : 2 ax2 - 2 6x2 - 6 ax + 6 6x - 8 a + 8 6 = 2(a — 6)x2 - 6(a - 6)x - 8(a - 6) = (a -6) (2x2 -6.x -8) = 2{a-6)(x2-3x-4) = 2 (a - 6) (X - 4) (X + 1). In this example the factor (a — 6) might have been replaced by a letter. 2. a2 + 62 - c2 - 9 - 2 a6 + 6 c. Solution : a2 + 62 - c2 - 9 - 2 a6 + 6c = a2 - 2 a6 + 62 - (c? - 6c + .9) = (a - 6)2 -^ (c - 3)2 . = (a - 6 + c - 3) (a - 6 - c + g. 3. (3x-2/)(2a+p)-(3x-2/)(a-9). 4. (4a-66)(3m-2p) + (a + 56)(3m-2p). 5. (7a-32/)(5c-2d)-(6a-22/)(5c-2d). 6. (X - 2/) (3a + 46) - (4a - 56) (X - 2/) - (X - y) (2a - 86). 22 ALGEBRA TO QUADRATICS 7. 6(x + 2/)2-ll(x4-2/)-7. 8. 4a2 - 12a6 + 962 _ x2 - 2x - 1. 9. x2a2 + 2 x2a + x2 - a2 - 2 a - 1. 10. ax2 + 6 ox + 9a - &x2 - 66x - 96. fll. 4 (a - 6)2 - 5(a2 - 62) - 21 (a + 6)2. 12. 6(x + ?/)2 - 12 (x2 -y2)^4{x- yf. 13. a262x2 - a262 - 2 a6x2 + 2 a6 + x2 - 1. 14. (x-2y)(2a-36)-(96-10)(x-2?/). / 35. Factoring binomials of the form d^ ± 6*^. By § 27, b/ ^^ V ci" + ^" = (« + ^'Xa"-^ - a"-2^ + a«-3^»2 _|_ ^,«-i>)^ where w is odd. One can factor by inspection any binomial of the given form by reference to these equations. EXERCISES Factor the following : 1. x^ — 2/6. Solution : x^ - ye - ^xB _ ^3) (a;3 + ^3) = (X2 + Xy + 7/2) (X - y) (X2 - Xy + 2/2) (x _|. y). 2. x6 + 125. 3. x^ - 1. 4. xi2 - 2/12. 5. x^ — 2/9. 6. Xl8 - 2/18. 7. a;16 _ yl6. 8. a2x3 + a5. 9. x* - a82/4. . 10. 216 a + a*. 11. ox* - 16 a. 12. 3a7 - 96 65a2. 1^-"" 13. 2,1 x^^ + 64 2/8. ^"^^ 14. 27 x52/7 + x22/4. «--^ 15. 16a468-81ci6d8,/^ 36. Highest common factor. An expression that is not further divisible into factors with rational coefficients is called prime. If two polynomials have the same expression as a factor, this expression is said to be their common factor. The product of the common prime factors of two polynomials is called their highest common factor, or H.C.F. The same common prime factor may occur more than once. Thus (x — 1) 2 (x + 1) and (x - 1)2 (a; - 2)2 have (x -X^ as their H.C.F. v-^— c^f FACTORING 23 37. H.C.F. of two polynomials. The process of finding the H.C.F. is performed as follows: EuLE. Factor the polynomials. The product of the common prime factors is their H.C.F. EXERCISES [l^'^-Xi^\)i^^^UH Find the H.C.F. of the following: 1. 4 ab^x^ - 8 ab^x^ + 4 ab^ and 6 abx"^ + 12 abx + 6ab. Solution : — 4 ab'^x^ - 8 ab^^ + 4 ab^ = 4a62(x4- 2x2+1) = 4a62(x-l)2(x + l)2. 6 a6x2 + 12 abx + 6ab = 6a6(x2 + 2x + 1) = 6a6(x + 1)2. The H.C.F. is then 2 a6(x + 1)2. 2. x^ — y^ and x2 — y"^. 3. x3 + x2 - 12 X and x2 + 5x + 4. 4. 9mx2 — Qmx + m and 9nx2 — n. 5. 6x- 4x2 + 2 ax -3a and 9 - 4x2. 6. 12 a2 - 3(3a6 + 27 62 and 8a2 - 1862. 7. 3 a2x - 6 a6x + 3 62x and 4 02^/ - 4 62y. 8. 2 X - 46 - x2 - 2 6x and 4x - 5x2 - 6. 9. 6x^-7 ax2 - 20 a2x and 3 x2 + ax - 4 a2. 38. Euclid's method of finding the H.C.F. When one is unable to factor the polynomials whose H.C.F. is sought, the problem may nevertheless be solved by use of a method which in essence dates from Euclid (300 e.g.). The validity of this process depends on the following Principle. If a polynomial has a certain factor, any multiple of it has the same factor. Let x» + ^x»-i + 5x»-2 + . . . +£: and . x« + ax'»-i +6x'»-2 H 1- ^ be represented by F and 6? respectively. The letters A, B, ■ ■, K and a, 6, • • •, Z represent integers, and m, the degree of G, is no greater than n, the degree of F. We seek a method of finding the H.C.F. of F and G if any exists. Call Q the quotient obtained by dividing F by G, and call B the remainder. Then (§ 26) ^' F=QG-^B, (1) 24 ALGEBRA TO QUADRATICS where the degree of E in x is not so great as that of G. Now whatever the H.C.F. of i^'and G may be, it must also be the H.C.F. of G and B. For since F-QG = B, the H.C.F. of F and G must be a factor of the left-hand member, and hence a factor of R, which is equal to that member. Also every factor common to G and R must be contained in F, for any factor of G and E is a factor of the right-hand member of (1), and hence of F. Thus our problem is reduced to finding the H. C. F. of G and R. Let Qi and Bi be respectively the quotient and remainder obtained in dividing G by R. Then (? = QiB 4- Ru where the degree of Ei in x is not as great as that of B. By reasonirg simi- lar to that just employed we see that the H.C.F. of G and R is also the H.C.F. of B and Bi. Continue this process of division. Let B = Q2R1 + R2, -Bi = QsRi + Eg. until, say in Rk = Qk+2Rk + \ + Bk+i, either B^ is exactly divisible by B^ +1 (i.e. E^- + 2 = 0), or i?;t + 2 does not con- tain X. This alternative must arise since the degrees in x of the successive remainders E, Ei, E2, • • • are continually diminishing, and hence either the remainder must finally vanish or cease to contain x. Suppose Bk + 2=0. Then the H.C.F. of Bk and Bk + 1 is Ea.- + 1 itself, which must, by the reasoning given above, be also the H.C.F. of F and G. If Rk + 2 does not contain x, then the H.C.F. of F and G, which must also be a factor of Ea + 2, can contain no x, and must therefore be a constant. Thus F and G have no common factor involving x. This process is valid if the coefficients of F and G are rational expressions in any letters other than x. 39. Method of finding the H.C.F. of two polynomials. The above dis- cussion we may express in the following Rule. Divide the polynomial of higher degree {if the degrees of the polyno- mials are unequal) by the other, and if there is a remainder, divide the divisor by it ; if there is a remainder in this process, divide the previous remainder by it, and so on until either there is no remainder or it does not contain the letter of arrangement. If tJiere is no remainder in the last division, the last divisor is the H. C. F. If the last remainder does not contain the letter of arrangement, then the polynomials have no common factor involving that letter. In the application of this rule any divisor or remainder may be multiplied or divided by any expression not involving the letter of arrangement without affect- ing the H.C.F. FACTORING ?6 EXERCISES Find the H.C.F. of the following: 2x 1. 2x* + 2x3-x2 Solution : /Q ^ Multiply by - 1 andx4 + x3 + 4x + 4. 2J2x* + 2x3-x2- 2x-l |xH x^+ 4x + 4 2x^ + 2x3 + 8x + 8 ■1, -x2-10x-9 *■ ^' ^ x^-\-10x+9\ x^+ x3+ 4x + 4 |x2-9x + 81 Ri x4 + 10x3+ 9x2 ,- [^-«) - 9x3- 9x2+ 4x ^t4H/l - 9x3-^90x2- 81x I Divide by - 725, (9-!. c , ec, x + 9| x2 + lOx + 9 |x + l \ 81x2+ 85X+4 81x2 + 810x+729 - 726ig 7 a 6 X2 + >y /'te.F. 9x + 9 9x + 9 Thus the H.C.F. is x + 1. This process may be performed in the following more compact form. 2 2x4 + 2x3-x2- 2x-l a^+ x^+ 4x + 4 x2-9x + 81 2x* + 2x3 + 8x + 8 -x2 - lOx-9 X4 + I0x3+ 9X2 -1 - 9x^- 9x2+ 4x x + 9 x2 + 10x + 9 - 9x3- 90x2- 81 X X2+ X 81x2+ 85x + 4 9x+9 9x + 9 81x2+810x +729 -725x -725 -725 X +1 Result: x + 1. 2. x2 + 6 X - 7 and x^ - 39x + 70. 3. x^ — X* — X + 1 and 5 x* — 4 x^ — 1. 4. x8 + 2x2 + 9 and - 6x3 - iia;2 + i5x + 9. 5. x3 - 2x2 - 15x + 36 and 3x2 - 4x - 15. 6. x* - 3x3 + x2 + 3x - 2 and 4 x3 - 9x2 + 2 X + 3. 7. 4x8 - 18x2 + 19x - 3 and 2x* - 12x3 + 19x2 _ 6x + 9. 8. X* + 4x3 _ 22x2 - 4x + 21 and x* + 10x8 + 20x2 - lOx - 21. 9. 6 a*x3 - 9 a3x2y - 10 a^ xy^ + 16 ay^ and 10 a^xV - 1^ «*«^y* + 8 a^*y^ 12 a2x3y6. 26 ALGEBRA TO QUADRATICS 40. Least common multiple. The least common multiple of two or more polynomials is the polynomial of least degree that con- tains them as factors. We may find the least common multiple of several polynomials by the following EuLE. Multiply together all the factors of the various poly- nomials, giving to each factor the greatest exponent with which it appears in any of the polynomials, 41. Second rule for finding the least common multiple. When only two polynomials are considered the previous rule is evidently equivalent to the following EuLE. Multiply the polynomials together and divide the product hy their highest common factor. EXERCISES Find the least common multiple of the following : 1. x2 - y2,a;2 + ay — ax - xy, and a;2 - 2 xy + y\ Solution : x^ -y'^ = {x - y)ix -^ y). x'^ + ay -ax-xy = {x- y) {x — a). x^-2xy + y^ = {x- yf. Thus the L.C.M. = (x - yY (x + y) (x - a). 2. 4a26c, 6a62, and 12 c2 3. 9x?/2, 6x22/3, and ZxyH"^. 4. (X + 1) (x2 - 1) and x3 - 1. 5. x* + 4x22/2 and x2 + 2 2/2 - 2 y. 6. 4x2 - 9?/ and 4x2 - 12x2/ + 9?/2. 7. x2 - 4 X + 3, x2 - 1, and x2 - ax — x + a. 8. x-1, 2x2-5x-3, and2x8-7x2 + 2x + 3. 9. x* - 9x2 + 26x - 24 and x^ - 10x2 + 31 x - 30. 10. 2x2 - 3x - 9, x2 - 6x + 9, and 3x2 _ 9x - 6x + 3&. CHAPTEE III FRACTIONS 42. General principles. The symbolic statements of the rules for the addition, subtraction, multiplication, and division of alge- braic fractions are the same as the statements of the correspond- ing operations on numerical fractions given in (2), (3), and (4), § 6. This is immediately evident if we keep in mind the fact that algebraic expressions are symbols for numbers and that if the letters are replaced by numbers, the algebraic fraction becomes a nunierical fraction. 43. Principle I. Both numerator and denominator of a frac- tion may he multiplied (or divided) hy the same expression with- out changing the value of the fraction. This follows from (5), § 6. 44. Principle II. If the signs of both numerator and denomi- nator of a fraction he changed, the sign of the fraction remains unchanged. This follows from Principle I, when we multiply both numerator and denominator by — 1 . 45. Principle III. If the sign of either numerator or denomi- nator (hut not hoth) he changed, the sign of the fraction is changed. This follows from (6), § 6. 46. Reduction. A fraction is said to be reduced to its lowest terms when its numerator and denominator have no common factor. We effect this reduction by the following Eule. Divide hoth numerator and denominator hy their highest common factor. 27 28 ALGEBRA TO QUADRATICS EXERCISES * Reduce the following to their lowest terms. 12 ax2 - 12 ab^ 1. 4ax^-Sabx-\-4:al» 12ax2-12a62 12 a {x - b) {x ■{■ b) 4 ax2 - 8 a6x + 4 ab^ Solution : H.C.F. =4a{x-b). a;S + 4x *^ a — 2ax — lOx + 5 g 6x2 -8ax +2a2 2 a26 + 2 a62 - 2 abc x2-a2 * ■ 3 6c2 - 3 62c - 3 abc ' g a2 + 62 _ c2 + 2 a& 21x8-9x2 + 7x - 3 xi8 - ai8* • a2 - 62 4. c2 + 2 ac' ' 3x8 + 15x2 + x + 5 * ^j^ 2x2+ 3x-9 j^2 X* - x8 - X + 1 j^„ x3 + 3ax2 + 3a2x + a8 x2-9 ' ' 2x4 -x3-2x + l' * a2 + 2ax + x2 47. Least common denominator of several fractions. We have the following Rule. Find the least common multiple of the various denomi- nators. Multiply both numerator and denominator of each fraction hy the expression which will make the new denominator the least common multiple of the denominators. EXERCISES Reduce the following to their least common denominator. - 2 8 ^ 2x-3 1. -> , and — -. X 2x-l 4x2-1 Solution : The L.C.M. of the denominators is x (4x2 — 1) . Thus the frac- tions are 2 (4 x2 - 1) 3x(2x+l) ^^ x(2x-3) x(4x2-l)' x(4x2-l) ' ^^ x(4x2-l)' " , and 3. , ■ , and 6 + a a2-62 2x-8 4x2 + 4x-16 4x2-26 FRACTIONS 29 A 1 1 ^ 1 4. , , and aj3 _ y3 aj4 _ y4 a;2 - y2 5. , — - — , and c 2x-\ X ^ 2x-3 6. — , , and x2 - 2 X + 1 x2 - 1 (x + 1)2 -a b . c 7. : , , and a + b-c a-\-b -^c d^ + 2ab + b^ - c"^ 48. Addition of fractions. This operation we perform as follows : EuLE. Reduce the fractions to he added to their least common denominator. Add the numerators for the numerator of the sum, and take the least common denominator for its denominator. 49. Subtraction of fractions. This operation we perform as follows : Rule. Reduce the fractions to their least common denomi- nator. Subtract the numerator of the subtrahend from that of the minuend for the numerator of the result, and take the least common denominator for its denominator. 50. Multiplication of fractions. This operation we perform as follows : Rule. Multiply the numerators together for the numerator of the product, and the denominators for its denominator. 51. Division of fractions. This operation we perform as follows : Rule. Invert the terms of the divisor and multiply by the dividend. a Remark. Since a fraction is a means of indicating division, t~^^ and - are two expressions for the same thing. d 30 ALGEBRA TO QUADRATICS EXERCISES ^ Perform the indicated operations and bring the results into their simplest forms. a + b a — b I I* — w a + b ' a + b a — b a + & Solution ; a 4- 6 a — b a ^b a + 6 a + b a — b a—b a+b a^-l^ {a + 6)2 + (a ■ -6)2 a2-62 (a + 6)2 _ (a . -6)2 a2-62 (a + 6)2 + (a - -6)2 a2-62 ( 2 a2 + 2 62 a2 + 62 4a6 2a6 iW - 1 2.V * 3 + f + i (a + 6)2 - (a - 6)2 ' 1 + 1 ' 2.V * • 1-|-t\ 5 2-V- g 3 + f + l 7 2 + f + f * i-(-W ■f + l + i.V '1-l + f" 8. ^_^L±^. 9. ^-i^-^. 10. -i- + 6 26 ab ac be a — b a + b 11.1+1+1. 12. '?^l^+!-^ 13. 20= a6c c c a-la2-l -, 3a;-l 2«-7 -- 2ic-l 2x-5 14. . 15. l-3x 7 a;-2a;-4 16.—?^ L_. 17. ^ 8 4x-4 6a; + 6 3x-9 6«-15 18 « + ^ ■ g'^ + 2 a6 - 62 2a-36 3a-26 ' a-6 ■ a2-62 " " 12 a "*" 16a 20 ?. ■ Q 21 «t? + ^c acZ-6c 15(x-l) 10(x + l) 2cd{c-d) 2cd{c + d) 22 a? + y a;-y 4zy ^^ a;2 + a;(a + 6) + a6 gg-gg ' x-y x-\-y x2-y2' ' x2 - x(a + 6) + a6' x^ - 62' 24. i5 + il + i?£-l^. 25. ^ « 6 a 14 a 36 a 16 a a2 - 9 a + 14 ' a* - 6 a - 14 26. Z^ + lU/^-l + lV 27 g(«-'g) a(a + x) \y8 a;/ \y2 y x) a2 + 2 ttx + x2 a2-2ax + xa* 28. lzi^.lz^./i + _±_V 29. -A_ + _^ + _A._l. 1 + 6 a + a« V 1-a/ (x - 1)« (x - 1)2 ^ x - 1 x FRACTIONS 31 ^ 1 30. a + --L.. 31. -^^i 32. ^. c + ^ (^Y-1 x + l + l e \h/ X 33. 1 34. 35. a + ^ 1 .1 d x-\ 4h cH x-S 9 a , h ^ db 1,1 a2 + 62 ^ o«a + 6 a-6 a2-62 i + x 1-x „„ a a2-62 OO. • Of. • So. ' 1 1 1111 a3 + 63 (a + &)2 {a-hY 1 - « 1 + ic a h 39 • a;-3y ^ x + 3y ^^ _J^ a; + 2 ■ x2-2a;y-15?/2 " x2-8xi/ + 15y2' ' 3(x + l) 3(-4-3x + x2) ^^' 1 ri'+ 2y. ;• X y + z -Q 2a-36 + 4 3a-46 + 5 a-1 6 8 12 43 l ^"^ -^y^ x2 - y2 \ / x + y X - y \ ■ \x2 - 2/2 a;2 4. 2/2/ • \^a; _ y x^yj' X- 1 y- 1 z -1 -^ 3xy2 x__ y g ' yz -{• zx — XV 11 1 X y « 45 / 2x + y 2y-x x2 \ x2 + y2 \ X + y X - y x2 - y2/ ■ x2 - y2 -^a — 36 4a — 6 5a + 3c a2 — 6c 2a 40. 1 1 • 6a 26 9c 2ac 6 . „ 6cd! cda 47. 1 ^ (a - 6) (a - c) (a - d) (6 - c) (6 -d^(b-a) dab ^ (c-6)(c-a)(c-d) a6c "^ {d -a){d- 6) (d - c) ' 48 1 a-26 8 3a-4c2 9 5c2_--_66 ' 6a 3a6 46 8ac2 8c2 126c2 ' X-1 X + 1 (X - 1)2 (X + 1)2 X2 - 1 (X2 - 1)2 CHAPTER IV EQUATIONS 52. Introduction. An equation is a statement of equality between two expressions. We assume the following Axiom. If equals he added tOy subtracted frorriy multiplied hy^ or divided hy equals, the results are equal. As always, we exclude division by zero. In dividing an equation by an alge- braic expression one must always note for what values of the letters the divisor vanishes and exclude those values from the discussion. 53. Identities and equations of condition. Equations are of two kinds : First. Equations that may be reduced to the equation 1 = 1 by performing the indicated operations are called identities. Thus 2 = 2, a -6= (3a -26) -(2a -6) are equations of this type. In identities the sign = is often replaced by =. It should be noted that identities are true whatever numerical values the letters may have. Second. Equations that cannot be reduced to the form 1 = 1, but which are true only when some of the letters have particular values, are called equations of condition or simply equations. Thus z=2 cannot further be simplified, and is true only when x has the value 2. Also a; = 2 a is true only when x has the value 2 a or a has the value - • If in this equation x is replaced by 2 a, the equation of condition reduces to an identity. The number or expression which on being substituted for a letter in an equation reduces it to an identity is said to satisfy the equation. Thus the number 5 satisfies the equation x^ — 24 = 1. The number 3 satisfies the equation (x — 3) (a; + 4) = 0. 32 EQUATIONS 33 The process of finding values that satisfy an equation is called solving the equation. The development of methods for the solu- tions of the various forms of equations is the most important question that algebra considers. In an equation in which there are two letters it may be possible to find a value which substituted for either will satisfy the equa- tion. Thus the equation x — 2 a = Ois satisfied if x is replaced by 2 -a, or if a is replaced by -• In the former case • we have solved for £c, that is, have found a value that substituted for x satisfies the equation. In the latter case we have solved for a. In any equation it is necessary to know which letter we seek to replace by a value that will satisfy the equation, that is, with respect to which letter we shall solve the equation. The letter with respect to which we solve an equation is called the variable. Values which substituted for the variable satisfy the equation are called roots or solutions of the equation. When only one letter, i.e. the variable, occurs in an equation, the root is a num- ber. When letters other than the variable occur, the root is expressed in terms of those letters. 54. Linear equations in one variable. An equation in which the variable occurs only to the first degree is called a linear equa- tion. To solve a linear equation in one variable we apply the following EuLE. Apply the axiom (§ 52) ^o obtain an equation in which the variable is alone on the left-hand side of the equation. The right-hand side is the desired solution. To test the accuracy of the work substitute the solution in the original equation and reduce to the identity 1=1. Since the result of adding two numbers is a definite number, and the same is true for the other operations used in finding the solution of a linear equation, it appears that every linear equation in one variable has one and only one root. When both sides of an equation have a common denominator, the numerators are equal to each other. This appears from multiplying both sides of the equation by the common denominator and then canceling it from both fractions. 34 ALGEBRA TO QUADRATICS EXERCISES Solve : - 4«-2 , 6a; Sx ^ ^•-^ + T = T + '- Solution : Transpose the term involving x, 4a;-2 5x 3x ^ ... . ^. 32x-16 + 25x-30a; ^ Add fractions, = 5. 40 Clear of fractions and simplify, 97 x — 216 x = 8. rt a(d^ + x^) , ax dx d d^ 4- x^ X Solution : Divide by a, = c + -. dx d Transpose the term involving x, d* + x2 X dx d = c. Add fractions, = c. dx Clear of fractions and simplify, , _ d X = -' c 3. {a-l)x = b-x. 4. {a-x){l-x) = z^-l. 5. a(x-a2) = 6(x-62). 6. 2a; - fx = f « - 1 - |x + 2. 7. 8x - 7 + X = 9x - 3 - 4x. 8. .617 x - .617 = 12.34 - 1.234x. 9. 3(2x-.3) = .6 + 5(x-.l). 10. 7 - 5x + 10 + 8x - 7 + 3x = x. 11. (x-3)(x-4) = (x-6)(x-2). 12. f {x\[|(|x+5)-10] + 3}-8 = 0. 13. (H-6x)2+(2 + 8x)2 = (H-10x)2. 14. 6 = 3x+i(x + 3)-^(llx-37). 15. 2(x + 5) (X + 2) = (2x + 7)(x + 3). 16. (7ix - 2|) - [4| -UH- 5a;)] = 18^. 17. 6x - 7(11 - X) + 11 = 4x- 3(20 -X). 18. (a - 6) (X - c) + (a + 6) (« + c) = 2(&x + ad). 19. 2x - 3(6 + f X) + ^(4 - X) - ^(3x - 16) = 0. 20. 5x-2 = fx + fx + fx + T'ffa; + H« + i|a;. 21. (a - 6) (a - c + x) + (a + &) (a + c - X) = 2 a*. 22. 12.9x - 1.46X - 3.29 - .99x - llx + .32 = 0. 23. 6.7x - 2^7.8 - 9.3x) = 5.38 - 4|(.28 + 3.6x). EQUATIONS S6 24.3-^ = ^-. 25. -^+l^c. 3 11 mx nx 3 X 26 x-J:^x-^^ 27 iil^^lll^?. x-S x-^' ' f (6x + l) 3' 28 t(a^-4) ^1 29 2x2-3x + 5 ^2 '1(3x4-5) 6* ■7x2-4x-2 7* 3o.» + l = 12 + l 31. i:i-%i = ^--l x2x9 J+x4^+x4 „rt« + &aj_c + cZx ««25x — 2_52x — 5 a + b ~ c + d' ' 3"7x-3~ 7 3x- 7* „- ox ex , /x , „- X + a & X - 6 a 34. — + — -^ — = h. 35. — = + -■ . bag b a a b 36. 5^ + i, = x-l. 3y 3x-19^5x-25^3 a X — 13 X + 7 3 12 1 3 2 38. L4-L^ = -l--L. 39. ^—^ + ^-^i^ + '-:z^ = o. 3 12 12 1 2'^x 3'^x S'x"^ ^^x-8x + 12 „, 18 ,-a(2x + l) Sox- 46 4 4U. 1- = z H • 41. — = — • x + 2 x-8 x + 2 36 66 6 .„ ax 6x 2a6 (a + 6)2x '4(6. i f- = < 6 a a + 6 ab 43. 8ix - - - 3|x- 4^x + 1 = 0. 6 .. 1 a + b 1 a — b 44. + = + a + b x a — b x 45 ^ (^ - ^) _L ^ + 8 _ 3(5x + 16) X- 7 X -4~ 5X-28 ' 46.^ a 6 ^^^ox-A_^ X b^c — X «c2 — X _ 5X-.4 1.3 -3x _ 1.8-8X 2 1.2 ^48. -^^ + -1^- + -^ = 2. 6x + 2 15x4-6 3x4-1 49.I^^2-i(. + 3) + 6 = ^-(^±?>. 3 6^ ' 2 36 ALGEBRA TO QUADRATICS 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 5x-l 3x + 2 ic2-30x + 2 3(x + l) 2(x-l) 6x2-6 3x-2 . 7x-3 . x + 100 = 10. x + 3 x + 2 x2 + 5x + 6 36(x-a) x-62 6(4a + cx) ^Q 5a 166 6a 16X-27 x + 3 _ 6 + 3x 4x-7 21 6 ~ 2 3 " a{b-x) b{c-x) _ a + 6 , /& , a\ 6x ex X \c &/ 5x-6 9-lOx 3x-4 3-4x 10 4-2x 3 3- 6 1.5x 7 4x2 6x-3 X-.5 3(2x-l) ^ + 6(2x-5) 2(2x-6) 3(2x-6) x« + i-3x»-i 3x«-i-x« x« „ 4x 4 2 ax — be 6x — ac_cx — 62 x — a x ab c^ be c a 3x X — 2a X — 26 X — 2c 6 + c a+c— 6 a+6— c a+6+c 2x« + 7x»-i 7x"-44x«-i 4X" + 27x'»-i 9 5X-14 3x + 3 /x + 1 18 a(x-3) 6(x-3) a2(x-l) 62(x-l) ^^ 6 a 62 "^ a2 ~ • (m + n)2x nx _ c 3nx mb 6 m (a — 6) 6 m (a — 6) 6 4(13x-.6) 3(1.2-x)_9x + .2 5 + 7x 6 "^ 2 -~20~"^~~r~"^'^- (^^±^(x-a) + ^^i:iA'(x-6) = 2a(2a + 6-x). 6 a ax — 6 ex — d (6n + dm)x + (6p + dg) _ a e mx—p nx — q (mx - p) (nx - g) ~" m n' EQUATIONS 37 55. Solution of problems. The essential step in solving a problem by algebra is the expression of the conditions of the problem by algebraic symbols. This is, in fact, nothing else than a translation of the problem from the English language into the language of algebra. The translation should be made as close as possible, clause by clause in most cases. In general the result sought should be represented by the variable, which for that reason is often called the unknown quantity. Example. What number is it whose third part exceeds its fourth part by sixteen ? Solution : " What number is it " is translated by x. Thus we let x represent the number sought. " Whose third part " is translated by - . " Exceeds its XX fourth part " is translated by , i.e. the third part less the fourth part leaves something. " By sixteen " gives us the amount of the remainder. Thus the translation of the problem into algebraic language is Let X represent the number sought. ^-^ = 16. 3 4 This equation should be solved and checked by the methods already given. PROBLEMS 1. What number is it whose third and fifth parts together make 88 ? 2. What number increased by 3 times itself and 5 times itself gives 99 ? 3. What is the number whose third, fourth, sixth, and eighth parts together are 3 less than the number itself? 4. What number is it whose double is 7 more than its fourth part ? 5. In 10 years a young man will be 3| times as old as his brother is now. The brother is 7| years old. How old is the young man ? 6. A father who is 53 years old is 3 years more than 12 1 times as old as his son. How old is the son ? 7. If you can tell how many apples I have in my basket, you may have 4 more than ^, or, what is the same thing, 4 less than i of them. How many have I? 8. If Mr. A received ^ more salary than at present, he would receive $2100. How much does he receive? 9. A boy spends ^ of his money in one store and \ of what remains in another, and has 24 cents left. How much had he ? 38 ALGEBRA TO QUADRATICS 10. A man who is 3 months past his fifty-fifth birthday is 4^ times as old as his son. How old is the son ? 11. In a school are four classes. In the first is ^ of all the pupils ; in the second, ^ ; in the third, j- ; in the fourth, 37. How many pupils are in the school ? 12. A merchant sold to successive customers ^, |, and ^ of the original length of a piece of cloth. He had left 2 yards less than half. How long was the piece ? 13. How may one divide 77 into two parts of which one is 2| times as great as the other ? 14. The sum of two numbers is 73 and their difference is 15. What are the numbers ? 15. A father is 4i times as old as his son. Father and son together are 27 years younger than the grandfather, who is 71 years old. How old are father and son ? 16. The sum of two numbers is 999. If one divides the first by 9 and the second by 6, the sum of these quotients is 138. What are the numbers ? 17. The first of two numbers whose sum is a is b times the second. What are the numbers ? 18. If the city of A had 14,400 more inhabitants, it would have 3 times as many as the city of B. Both A and B have together 12,800 more than the city of C, where there are 172,800 inhabitants. How many are in A and B ? 19. Two men who are 26 miles apart walk toward each other at the rates of 3| and 4 miles an hour respectively. After how long do they meet ? 20. A courier leaves a town riding at the rate of 6 miles an hour. Seven hours later a second courier follows him at the rate of 10 miles an hour. How soon is the first overtaken ? 21. A can copy 14 sheets of manuscript a day. When he had been work- ing 6 days, B began, copying 18 sheets daily. How many sheets had each written when B had finished as many as A ? 22. The pendulum of a clock swings 387 times in 5 minutes, while that of a second clock swings 341 times in 3 minutes. After how long will the second have swung 1632 times more than the first? 23. The difference in the squares of two numbers is 221. Their sum is 17. What are the numbers ? 24. If a book had 236 more pages it would have as many over 400 pages as it now lacks of that number. How many pages has the book ? 25. A man is now 63 years old and his son 21. When was the father 19 times as old as his son ? EQUATIONS 39 ' 26. If 7 oranges cost as much less than 50 cents as 13 do more than 50 cents, how much do they cost apiece? 27.. The numerator of a fraction is 6 less than the denominator. Dimin- ish both numerator and denominator by 1 and the fraction equals |. Find the fraction. 28. The sum of three numbers is 100. The first and second are respec- tively 9 and 7 greater than the third. What are the numbers ? 29. Out of 19 people there were f as many children as women, and 1^ times as many men as women. How many were there of each ? 30. A boy has twice as many brothers as sisters. His sister has 5 times as many brothers as sisters. How many sons and daughters were there ? ySl. A dealer has 5000 gallons of alcohol which is 85% pure. . He /wishes to add water so that it will be 75% pure. How much water must he add? i V 32. How much water must be added to 5 quarts of acid which is 10% full strength to make the mixture 8|% full strength ? .^33. A merchant estimated that his supply of coffee would last 12 weeks. He sold on the average 18 pounds a week more than he expected, and it lasted him 10 weeks. How much did he have ? 34. At what time between 3 and 4 o'clock are the hands of a clock point- ing in the same direction ? 35. At what time between 11 and 12 o'clock is the minute hand at right angles to the hour hand ? ^^6. A merchant bought cloth for |2 a yard, which he was obliged to sell for $1.75 a yard. Since the piece contained 3 yards more than he expected, he lost only 2%. How many yards actually in the piece ? 37. A man has three casks. If he fills the second out of the first, the latter is still f full. If he fills the third out of the second, the latter is still I full. The second and third together hold 100 quarts less than the first. How much does each hold ? 38. A crew that can cover 4 miles in 20 minutes if the water is still, can row a mile downstream in f the time that it can row the mile upstream. How rapid is the stream ? 39. A cask is emptied by three taps, the first of which could empty it in 20 minutes, the second in 30 minutes, the third in 35 minutes. How long is required for all three to empty the cask ? 40. A can dig a trench in f the time that B can ; B can dig it in f the time that C can ; and A and C can dig it in 8 days. How long is required by all working together ? 40 ALGEBRA TO QUADRATICS 56. Linear equations in two variables. A simple equation in one variable has one and only one solution, as we have abeady seen (p. 33). On the other hand, an equation of the first degree in two variables has many solutions. For example, Sx + 7y = l is satisfied by innumerable pairs of numbers which may be sub- stituted for X and y. For, transposing the term in y, we get X- 3 , from which it appears that when y has any particular numerical value the equation becomes a linear equation in x alone, and hence has a solution. Thus, when y = 1, ic = — 2, and this pair of values is a solution of the equation. Similarly, cc = — 9, y = 4 also satisfy the equation. 57. Solution of a pair of equations. If in solving the equation just considered, the values of x and y that one may use are no longer unrestricted in range, but must also satisfy a second linear equation, we get usually only a single pair of solutions. Thus if we seek a solution, that is, a pair of values of x and y satisfying Sx-\-7y = l, such that also x + y = -ly we find that the pair of values x = — 2, y = 1 satisfy both equa- tions. Any other solution of the first equation, as, for instance, x=—9, y = 4, does not obey the condition imposed by the second. Two equations which are not reducible to the same form are called independent. Thus 6a; -82/ -4 = and 3 a; — 4 y = 2 are not independent, since the first is readily reduced to the second by transposing and dividing by 2. They are, in fact, essentially the same equation. On the other hand, . „ ' X — 4y=2 and 3 X — 4 y = 2 are not reducible to the same form and are independent. Since dependent equa- tions are identical except for the arrangement of terms and some constant factor, all their solutions are common to each other. = -% EQUATIONS 41 This principle we may state as follows : Two equations , i, , r. ax + oy + c = and a'% + &'?/ + C = are dependent when and only when £_ &^ _ c Independent equations in more than one variable which have a common solution are called simultaneous equations. Two pairs of simultaneous equations which are satisfied by the same pair (or pairs) of values of x and y and only these are called equivalent. Thus r3.+7a==l. ^^ ra==- are equivalent pairs of equations. 58. Independent equations. We now prove the following Theorem. If A=0 and B = represent two in equations^ then the pairs of equations ^ = ^'(1) and {'''■^f^'!: (2) are equivalent where a, b, c, and d are any numbers such that ad — be is not equal to zero. The letters A and B symbolize linear expressions in x and y. Evidently any pair of values of x and y that makes both A = and B = 0, i.e. satisfies (1), also makes aA-{-bB=0 and cA + dB = 0, i.e. also satisfies (2). We must also show that any values of x and y that satisfy (2) also satisfy (1). For a certain pair of values of x and y let aA-{-hB = 0, (3) cA + dB= 0. (4) Multiply (3) by c and (4) by a (§ 52). Then acA + bcB = 0, (5) acA + adB = 0. (6) 42 ALGEBRA TO QUADRATICS Subtract (5) from (6) (§ 52), (ad-bc)B = 0. Thus, by § 5, either ad — be = ot B = 0. But ad — be is not zero, by hypothesis ; consequently -6 = 0. Similarly we could show that ^ = 0. Thus if we seek the solution of a pair of equations ^ = 0, jB = 0, we may obtain by use of this theorem a pair of equiva- lent equations whose solution is evident, and find immediately the solution of the original equations. 59. Solution of a pair of simultaneous linear equations. The foregoing theorem affords the following KuLE. Multijply each of the equations by some number such that the coejfficunts of one of the variables in the resulting pair of equations are identical. Subtract one equation from the other and solve the resulting simple equation in one variable. Find the value of the other variable by substituting the value just found in one of the original equations. Check the result by substituting the values found for both variables in the other equation. Example. Solve 3x-l-7y = l, (1) x + y=-\. (2) Solution : Multiply (1) by 1 and (2) by 3, 3x + 7?/ = l, 3x + 32/ = -3 Subtract, 4y = 4 Substitute in (2), x + l = 1. -1. Check: Substitute in (1), « = -2. 3. (- - 2) + 7 . 1 = -6 + 7 60. Incompatible equations. Equations in more than one vari- able that do not have any common solution are called incompatible. EQUATIONS 43 Theokem. The equations ax-\-by = c, ^ (1) afx-\-b'i/ = c' '*(2) are incompatible when and only when aV — ba^ — 0. Apply the rule of § 59 to find the solution of these equations. Multiply (1) by a' and (2) by a. We obtain aa^x + a^by — ca\ aa'x -f- ab'y = ac'. Subtract, (ab' — a'b) y = ac' — ca'. If now ab' — a'b is not zero, we get a value of y ; but since under our hypothesis ab' — a'b = 0, we can get no value for y since divi- sion by zero is ruled out (§ 7). Thus no solution of (1) and (2) exists. Example. Solve Sx + 1y = l, (1) 6x + 14y = l. (2) Solution : Multiply (1) by 2, Qx + Uy = 2 6x+ Uy= 1 Subtract, 0=1 which is absurd. Thus no solution exists. 61. Resum^. We observe that pairs of equations of the form ax -\- by -{- c = Oj a'x -f J'y + c' = fall into three classes : (a) Dependent equations, which have innumerable common solutions. ^ ^ ^ Tlien a'^b'^7'' W (b) Incompatible equations, which have no common solution. Then ^^, _ ^,j ^ ^^ ^^^ ^^^ -^ ^^^ ^^^^^ (c) Simultaneous equations, which have one and only one pair of solutions. Then ab' - a'b ^ 0. 18 44 ALGEBRA TO QUADRATICS EXERCISES Solve and check the following : . 2x + 6y = l, 2 4x-6y,= 8, , 6x + 8y ' Qx + 1y = S. ' |x-2/ = f. •x + |y = 3. ^ 7x-3y = 27, ^ 2x-|2/ = 4, ^ |y = ^x-l, * 6x-6y = 0. ■3x-|2/=:0. '^-^yrzfx-l, „ 5x-4y + l = 0, g 3x + 4y = 253, g 6x + 3y + 2 = 0, 1.7x-2.22/ + 7.9 = 0. 'y = 5x. ■3x + 2y + l=0. j^Q x + my = a, j^j^ x + y = |(5a + 6), ' x — ny = b. ' X — y = |(a + 6 6). -2 2x-3y = -5a, -, |x-i(y + l) = l, • 3x-22/ = -5&. • i(a; + l) + |(y-l) = 9. -- 3.5x + 2i2/=13 + 4fx-3.5y, -- 3x + 2 ?/ = 5a2 + a6 + 5&2^ ' 2ix + .8y = 22^ + .7x-3i2/. " 3?/ + 2x = 6a2 - a6 + 662. 16. 3 8 x+r 16 _ 4 X 2/ = 3, = 4. Hint. Ketain fractions. 18. X - c y-C a 6' r x-y = a - h. n/\ x + 2y 2x-2/ + 1 + 1 = 2, 17. 1 1_6 - 4- - — -1 X y 6 111 X y Q 3x + l_4 19. 4-2y 3 X + y = 1. - + ^ = c, 20. r"^"^; 21. "^ ^ ^^-^ + ^ = 5. ■ ^ + ^ = ci. X — 2/ + 3 «! 6i 5 7 22. 24. x + 2y 2x + 2/ 7 ^ 6 3x-2~6-2/* .9x-.7y + 7.3 _ 13X-152/ + 17" 1.2x-.2y + 8.9 ^ ^ 13x-16y + 17 X y 1 .2. a + 6 a — 6 a — 6 X _ y _ 1 a + 6 a — 6 a + 6 x+1 a+6+c 2.3 y + l~a-6 + c' X — 1 a + 6 — c y-1 a-6— c 25. y _6 27. a- a - c a2 - X a^-y _ EQUATIONS 46 ^^^ y = 4 - 3x + x2. * • (4a; - 7) (X - 3) = y. „Q4Vx-3Vy = 6, g-xVa-y\/6 = a + 6, 3 Vx — 4 Vy = 1. , X -\-y = 2 Va. 22 xV2 + ?/V3 = 3V3, 32 4V^T7-5Vir^ = 7, 'xV3-y\/2 = 2V2. "3 VxTT - 7 Vy^ = 2. «^ Vx Vy „> Vx - 3 Vy + 3 12, ^^4 9 , —= + —= = 1. , + , = 4. Vx Vy Vx - 3 V 2/ + 3 . ,, , a + 6 + 1 x + ly + 2 2(x-y) (a_&)[x + (a+&)y]=a-6+l. 3(x - 3)-4(y- 3) = 12(2y -x> 62. Solutions of problems involving two unknowns. Tlie same principle of translation of the problem into algebraic symbols should be followed here as in the solution of problems leading to simple equations (p. 37). PROBLEMS 1. The difference between two numbers is 3|. Their sum is 9|. What are the numbers? 2. What are the numbers whose sum is a and whose difference is 6 ? 3. A man bought a pig and a cow for $100. If he had given $10 more for the pig and $20 less for the cow, they would have cost him equal amounts. What did he pay for each ? 4. Two baskets contain apples. There are 51 more in the first basket than in the second. But if there were 3 times as many in the first and 7 times as many in the second, there would be only 6 more in the first than in the second. How many apples are there in each basket? 5. A says to B, "Give me $49 and we shall then have equal amounts." B replied, "If you give me $49, I shall have 3 times as much as you. How much had each? 6. A man had a silver and a gold watch and two chains, the value of the chains being $9 and $25. The gold watch and the better chain are together twice and a half as valuable as the silver watch and cheaper chain. The gold watch and cheaper chain are worth $2 more than the silver watch and the better chain. What is the value of each watch ? 46 ALGEBRA TO QUADRATICS 7. What fraction is changed into ^ when both numerator and denomi- nator are diminished by 7, and into its reciprocal when the numerator is increased by 12 and the. denominator decreased by 12 ? 8. A man bought 2 carriage horses and 5 work horses, paying in all $1200. If he had paid $5 more for each work horse, a carriage horse would have been only J more expensive than a work horse. How much did each cost? 9. A man's money at interest yields him $540 yearly. If he had received \% more interest, he would have had $60 more income. How much money has he at interest ? 10. A man has two sums of money at interest, one at 4%, the other at 5%. Together they yield $750. If both yielded 1% more interest, he would have $165 more income. How large are the sums of money ? 11. A man has two sums of money at interest, the first at 4%, the second at 3^%. The first yields as much in 21 months as the second does in 18 months. If he should receive \% less from the first and \% more from the second, he would receive yearly $7 more interest from both sums. What are the sums at interest? 12. What values have a mark and a ruble in our money if 38 rubles are worth 14 cents less than 75 marks, and if a dollar and a ruble together make %\ marks ? 13. A chemist has two kinds of acid. He finds that 23 parts of one kind mixed with 47 parts of the other give an acid of 84 1% strength and that 43 parts of the first with 17 parts of the second give an 80f% pure mixture. What per cent pure are the two acids ? 14. Two cities are 30 miles apart. If A leaves one city 2 hours earlier than B leaves the other, they meet 1\ hours after B starts. Had B started 2 hours earlier, they would have met 3 hours after he started. How many miles per hour do they walk ? 15. The crown of Hiero of Syracuse, which was part gold and part silver, weighed 20 pounds, and lost 1^ pounds when weighed in water. How much gold and how much silver did it contain if 19;^ pounds of gold and 10^ pounds of silver each lose one pound in water? 16. Two numbers which are written with the same two digits differ by 36. If we add to the lesser the sum of its tens digit and 4 times its units digit, we obtain 100. What are the numbers ? 17. A company of 14 persons, men and women, spend $48. If each man spends $4 and each woman $3, how many men and how many women are in the company? EQUATIONS 47 63. Solution of linear equations in several variables. This process is performed as follows : KuLE. Eliminate one variable from the equations taken in pairs, thus giving a system of one less equation than at first in one less variable. Continue the process until the value of one variable is found. The remaining variables may be found by substitution. Special cases occur, as in the case of two variables, where an infinite number of solutions or no solutions exist. Where no solution exists one is led to a self- contradictory equation on application of the rule. See exercise 17, p. 48. EXERCISES Solve and check the following : « + 2/ + 2 = 9, 1. a; + 2y + 42 = 16, x + 3y + 9z = 23. Solution : x + y -[■ z = Q x -{■ y ■}- z— ^ x + 2y -{-iz = 15 x + 3y-f 9z=:23 2/ + 32= 6 2y + 8z = U y + 4z= 7 y + Sz = 6 y -\-4z = 7 2=1 y + 3 = 6. y = 3. a; + 3 + 1 = 9. x = S. Check : 5 + 9 + 9 = 23. X + y = 37, x + y = xy, 2.x + 2 =25, 3. 2x + 2z=x2;, y + z = 22. Sz + Sy = zy. Hint. Divide the equations by xy, xz, yz respectively. X + y 4- z = 17, X + y + z = 36, 4. X + z - y = 13, 5. 4x = 3y, x + 2;-2y = 7. 2x = 3z. L3x-1.9y=.l, 2x+2y + « = a, 6. 1.7y-l.lz = .2, 7. 2y-i-2z+x = 6, 2.9x-2.1z = .3. 2z+2x + y = c. 48 ALGEBRA TO QUADRATICS x + 2y = 6, 8. y + 22; =8, z + 2u = ll, u + 2x = Q. y z 10. i + l = 25, X z 1 + 1 = 20. X y "y _oo iy-Zx--''' 12. . ^' -15, 9. X + y = m, y + z = a, z + u = n, u — x = b. 11. xy _1 ; + y~5' xz 1 x + z 6 yz _ 1 y-{-z~ 1 y + 1 13.?^ = 4, X-32 ' z + 1 yz ^^2. 2; + 3_l 4y-5z x + 1 2 14. 312/ = X + 2 + 12, 15. X + 2 = 2| 2/ - 14, 4^ 2 = X + y + 16. y + 2; = 3f X - 32. x + 2y-z=4.6, x + 2y + 32 = 15, y + 22;-x = 10.1, 17. 3x + 5y + 7z = 37, 2 + 2x-y = 6.7. 5x + By + llz = 59. 7x + 6y + 72 = 100, (x+2)(2y + l) = (2x+7)y, x_2y + 2 = 0, 19. (x-2)(3z + l) = (x+3)(32-] 3x + y-2« = 0. (l+l)(«+2) = (y+3)(z+l). CHAPTEE V RATIO AND PROPORTION 64. Ratio. The ratio of one of two numbers to the other is the result of dividing one of them by the other. a The ratio of a to 6 is denoted by a : 6 or by - • The dividend in this implied division is called the antecedent, the divisor is called the consequent. 65. Proportion. Four numbers, a, b, c, d, are in proportion when the ratio of the first pair equals the ratio of the second pair. This is denoted by a : 6 = c : d or by - = - • The letters a and d are called the extremes, b and c the means, of the proportion. 66. Theorems concerning proportion. If a, b, c, d are in pro- portion, that is, if a:b = G:doT- = -j (I) b d ^ ^ then ad = be, (II) b:a = d:c, ' (III) a:c = b:d, (IV) a -\- b : a = c -{- d : Cf (V) a — b : a = c — d : Cj (VI) a-{-b:a — b=:c-{-d:c — d: C^II) Equation (III) is said to be derived from (I) by inversion. Equation (IV) is said to be derived from (I) by alternation. Equation (V) is said to be derived from (I) by composition. Equation (VI) is said to be derived from (I) by division. Equation (VII) is said to be derived from (I) by composition and division. 49 50 ALGEBRA TO QUADRATICS 67. Theorem. If a numher of ratios are equal, the sum of any number of antecedents is to any antecedent as the sum of the corresponding consequents is to the corresponding consequent. Let a\h — c\d = e\f= g '.h^ or a c e g 'h~d~f~h' To prove a + c + e b-hdi-f g h If a c e g 1~'d~f~h~''' we have a = hr, c = dr, g = hr. Divide the sum of the first three equations by the last and we get gj^c + e ^ b + d-\-f g h 68. Mean proportion. The mean proportional between two num- bers a and c is the number b, such that a:b =^b : c. By (II), § 66, we see that ac = b^. EXERCISES If a : 6 = c : d, prove that : a2 . , c2 1. a + h: = c + d: a + b Solution : By (V), § 66, Squaring, we get a + b c + d a -hb _c + d a c (g + b)^ ^ (c + d)^ a2 c2 ' a + b _ c + d ^^ a2 ~ c2 ' a + 6 c + d a + b: - = c + d a + b c-\- d I RATIO AND PROPORTION 61 2. a2 : 62 = c2 : d2. 3. a + b:c + d = a:c. 4. ma:mh = nc: nd. 5. a2 : c2 = a2 + 62 . c2 + d^. 6. a2 + 62 : _^ = c2 + d52 : _^. 7. Va2 + c^ : VPTd^ = a : 6. a + 6 c + cZ 8. ma + n6 : ra + s6 = mc + n(i : re + sd. 9. a + h + c + d:a — b-\-c — d = a + b — c — d:a — h — c + d. 10. Find the mean proportional between a^ + c^ and 62 + ^2. 11. Find the mean proportional between a2 + 62 + c^ and 62 + c2 + d\ Solve the following f or x : 12. 20:96 = x: 57. 14. x — ax:Vx = Vx : x. - „ Vx + 7 + Vx 4 + Vx 10. — — ^;;^ = V^rp^ -Vx4-v^ a-6a6 ac Hint. Use composition and division. 18. (?^^^ + „i)..^±^-„i = (a + bf:z. \a — I a + 6 13. 8 a6 : X : = 6c : 1| ac. 15. \--/x l-3Vx = = 1 4. 17. a + & fl ^2-62 = X : a - -6 CHAPTER VI IRRATIONAL NUMBERS AND RADICALS 69. Existence of irrational numbers. We have seen that in order to solve any linear equation or set of linear equations with rational coefficients we need to make use only of the operations of addition, subtraction, multiplication, and division. When, however, we attempt to solve the equation of the second degree, x^ = 2, we find that there is no rational number that satisfies it. Assumption. A factor of one memher of an identity between integers is also a factor of the other memher. Thus let 2 . a = 6, where a and b are integers. Then since 2 is a factor of the left-hand member, it must also be contained in 6. Theorem. iVb rational number satisfies the equation x^ = ^. Suppose the rational number - be a fraction reduced to i' lowest terms which satisfies the equation. Then or a2 = 2 b\ (1) Thus, by the assumption, 2 is contained in a^j and hence in a. Suppose a = 2 a'. Then by (1) 4 a'^ = 2 b% or 2 a'^ = h\ that is, 2 must also be contained in ft, which contradicts the hypothesis that t is a fraction reduced to its lowest terms. The fact that the equation x^ = 2 has no rational solution is analogous to the geometrical fact that the hypotenuse of an isosceles right triangle is incommensurable with a leg. 62 IRRATIONAL NUMBERS AND RADICALS 63 70. The practical necessity for irrational numbers. For tlie practical purposes of the draughtsman, the surveyor, or the machinist, the introduction of this irrational number is superflu- ous, as no measuring rule can be made exact enough to distin- guish between a length represented by a rational number and one that cannot be so represented. As the draughtsman does not use a mathematically perfect triangle, but one of rubber or wood, it is impossible to see in the fact of geometrical incommensurability just noted a practical demand from everyday life for the intro- duction of the irrational number. In fact the irrational number is a mathematical necessity, not a necessity for the laboratory or draughting room, as are the fraction and the negative number. We need irrational numbers because we cannot solve all quad- ratic equations without them, and the practical utility of those nimibers comes only through the immense gain in mathematical power which they bring. 71. Extraction of square root of polynomials. This process, from which a method of extracting the square root of numbers is immediately deduced, may be performed as follows : EuLE. Arrange the terms of the polynomial according to the powers of some letter. Extract the square root of the first term, write the result as the first term of the root, and subtract its square from the given polynomial. Divide the first term of the remainder hy twice the root already found, and add this quotient to the root and also to the trial divisor, thus forming the complete divisor. Multiply the complete divisor hy the last term of the root and subtract the product from the last remainder. If terms of the given polynomial still remain, find the next term of the root hy dividing the first term of the remainder hy twice the first term of the root, form the complete divisor, and proceed as before until the desired number of terms of the root have been found. 54 ALGEBRA TO QUADRATICS EXERCISES Extract the square root of the following : 1. a* - 2 a^x + 3 aH^ - 2 ax^ + x\ Solution : a* - 2 a^x + 3 a'^x'^ - 2 ax^ + x* \a^ — ax + x^ a* 2a2-ax| - 2 a^x + 3 aH^ - 2 ax^ + x* - 2 g^x + a^x^ 2a^-2ax + x^\ 2 a^^ - 2 ox^ + x* 2 g^x^ - 2 ax8 + x^ 2. 1 + x. 3. 1-x. 4. 3x2 _ 2x + x* - 2x3 + 1. 5. x* - 6x3 + 13x2 - 12x + 4. 6. x* + 2/4 + 2x3?/ - 2x2/3 _ a;2?/2. 7. 9x* - 12x3 + 34x2 - 20x + 25. 8. 49g4-42g36+37g262_i2a63+464. 9. 2g6- 2gc - 2&c H g2 + 62 4. c2. 10. W*102 + v*U^ + U>4v2 + 2 W3u2|0 _|_ 2 t>8t«2w + 2 W?3m2u. 72. Extraction of square root of numbers. We have the following Rule. Separate the mcmher into periods of two figures each, heginning o.t the decimal point Find the greatest number whose sqyiare is contained in the left-hand period. This is the first figure of the required root. Subtract its square from the first period, and to the remainder annex the next period of the number. Divide this remainder, omitting the right-hand digit, by twice the root already found, and annex the quotient to both root and divisor, thus forming the complete divisor. Multiply the complete divisor by the last digit of the root, subtract the result from the dividend, and annex to the remainder the next period for a new dividend. Double the whole root now found for a new divisor and pro- ceed as before until the desired number of digits in the root have been found. In applying this rule it often happens that the product of the complete divisor and the last digit of the root is larger than the dividend. In such a case we must diminish the last figure of the root by unity until we obtain a product which is not greater than the dividend. IRRATIONAL NUMBERS AND RADICALS 55 At any point in the process of extracting the square root of a number before the exact square root is found, the square of the result already obtained is less than the original number. If the last digit of the result be replaced by the next higher one, the square of this number is greater than the original number. There are always two values of the square root of any number. Thus Vi = + 2 or — 2, since (+ 2)2 = (— 2)2= 4. The positive root of any positive number or expression is called the principal root. When no sign is written before the radical, the principal root is assumed. EXERCISES Extract the square root of the following : 1. 2.0000. Solution : 2'.00'00'00'|1.414 1 2.4|1.00 96 281 1 400 281 2.824 1 11900 -A 11296 604 2. 96481. 3. 56169. 4. 3. 5. 877969. 6. 2949.5761. 7. 5. 8. 257049. 9. .00070128. 10. 99. 11. 69.8896. 12. .0009979281. 13. 12. 14. 49533444. 15. 9820.611801. 16. 160. 73. Approximation of irrational numbers. In the preceding process of extracting the square root of 2 we never can obtain a number whose square is exactly 2, for we have seen that such a number expressed as a rational (i.e. as .a decimal) fraction does not exist. But as we proceed we get a number whose square differs less and less from 2. Thus 1.2 = 1, less than 2 by 1. 1.42 = 1.96, less than 2 by .04. 1.41^ = 1.9881, less than 2 by .0119. 1.4142 = 1.999396, less than 2 by .000604. 5Q ALGEBRA TO QUADRATICS Though we cannot say that 1.414 is the square root of 2, we may say that 1.414 is the square root of 2 correct to three decimal places, meaning that (1.414)2 < 2 < (1.415)2. 74. Sequences. The exact value of the square root of most numbers, as, for instance, 2, 3, 5, cannot be found exactly in deci- mal form and so are usually expressed symbolically. By means of the process of extracting square root, however, we can find a number whose square is as near the given number as we may desire. We may, in fact, assert that the succession or sequence of numbers obtained by the process of extracting the square root of a number defines the square root of that number. Thus the sequence of numbers (1, 1.4, 1.41, 1.414, • • •) defines the square root of 2. 75. Operations on irrational numbers. Just as we defined the laws of operation on the fraction and negative nimibers (pp. 2-4), we should now define the meaning of the sum, difference, prod- uct, and quotient of the numbers defined by the seqilence of num- bers obtained by the square-root process. To define and explain completely the operations on irrational numbers is beyond the scope of this chapter. It turns out, however, that the number defined by a sequence is the limiting value of the rational num- bers that constitute that sequence, that is, it is a value from which every number in the sequence beyond a certain point differs by as little as we please. We may, however, make the following state- ment- regarding the multiplication of irrational nimibers : In the sequence defining the square root of 2, namely, (1, 1.4, 1.41, 1.414, • • •) we saw that we could obtain a number very nearly equal to 2 by multiplying 1.414 by itself. In general, we multi- ply numbers defined by sequences by multiplying the elements of these sequences; the new sequence^ consisting of the products, defines the product of the original numbers. Thus (1, 1.4, 1.41, 1.414, • • ) (1, 1.4, 1.41, 1.414, • • •) = (1, 1.96, 1.9881, 1.999396, • • •). The numbers in this sequence approach 2 as a limit, and hence the sequence may be said to represent 2, IRRATIONAL NUMBERS AND RADICALS 57 76. Notation. We denote the square root of a (where a repre- sents any number or expression) symbolically by Va, and assert or, more generally, ^ _ , Va • V^ = Va • b. Similarly, Va -^ V^ = Va -^ b. EXERCISES 1. Form five elements of a sequence defining VS. 2. Form five elements of a sequence defining V6. 3. Form five elements of a sequence defining V6. 4. Form, in accordance with the rule just given, four elements of the sequence \/2 • VS. Compare the result with the elements obtained in Ex. 3. 5. Form similarly the first four elements of product V2 • V6 with the first four elements obtained by extracting the square root of 10. 77. Other irrational numbers. The cube root and higher roots of numbers could also be found by processes analogous to the method employed in finding the square root, but as they are almost never used practically, they will not be included here. It should be kept in mind, however, that by these processes sequences of numbers may be derived that define the various roots of numbers precisely as the sequences derived in the pre- ceding paragraphs define the square root of numbers. The Tith root of any expression a is symbolized by Va. Here n is sometimes called the index of the radical. The principle for the multiplication and division of radicals with any integral index is given by the following Assumption. The product (or quotient) of the nth root of two numbers is equal to the nth root of the product (or quotient) of the members. Symbolically expressed, Va • -y/b = VoT^, Va H- V^ = Va -hb. 68 ALGEBRA TO QUADRATICS 78. Reduction of a radical to its simplest form. A radical is in its simplest form when the expression under the sign is integral (§ 11) and contains no factor raised to a power which equals the index of the radical ; in other words, when no factor can be removed from under the radical sign and still leave an integral expression. We may reduce a quadratic radical to its simplest form by the following Rule. If the expression under the radical sign is fractional^ multiply both numerator and denominator by some expression that will make the denominator a perfect square. Factor the expression under the radical into two factors^ one of which is the greatest square factor that it contains. ■ Take the square root of the factor that is a perfect square, and express the multiplication of the result by the remaining factor under the radical sign. If the radical is of the nth index, the denominator must be made a perfect nth power, and any factor that is to be taken from under the radical sign must also be a perfect nth power. EXERCISES Reduce to simplest form : 1. Vv. Solution : /T2 /12 . 5 /4 • 15 \5=\ 25 =\ 25 ~- = |Vl6. 5 2. vi- 3. V32. 4. Vf. 5. V27. 6. VA. 7. V243. 8. V'250. 9. Vi + 4. 10. Vi-|. 11. 8V75. 12. tV2762. 15. V¥ + |. 13. lV80a;8y*. 14. VtV + ^V 16. v-V^ + ^V "■ xS- ■ ^^ S ■'■ xlf ■ "M -->/!• «■f^/S■• -# -• ^^m■ - -Vi^- IRRATIONAL NUMBERS AND RADICALS 69 26. Vx^ - 2 x2y + X2/2. 27. V6x8-20x2 + 20x. 28 M^-^«'^ + « 29 / 2a;«-I2x2:+l8x „Q / 2 gs - 8 a2 + 8 g „- j a^ + a% -~ab' ' \8x-8x2 + 2x3' * \ 9(g-&.) g62 _ 63 79. Addition and subtraction of radicals. Radicals that are of the same index and have the same expression under the radical sign are similar. Only' similar radicals can be united into one term by addition and subtraction. We add radical expressions by the following Rule. Reduce the radicals to he added to their simplest form. , Add the coefficients of similar radicals and prefix this sum as the coefficient of the corresponding radical in the result. A rule precisely similar is followed in subtracting radical expressions. EXERCISES Add the following : 1. V27, \/48, and V75. Solution : V27 = - n/ 9 • 3 = 3 V3 V48=- v/I6.3= 4V3 \/75=^ v/25.3= 6V3 Sum = 12 \/3 2. V3 + 2V3. 3. 8V7-3V7. 4. ay/x-hy/x. 5. g + 2Vg + 3\/a + 2 Vl6g - \/27g. 6. 3 V8 + 4 V^ ~ 5 V50 - 7 V72 + 6 V98. 7. 8Vg+5\^-7Vg + 4Vg-6Vx-3 Vg. 8. 7 V4x + 4 Vox + 3 ViSx - 5 V36x - 2 V80x. 9. Vg^^ + Vl6g - 106 + Vgx2 - 6x2 _ V9(g - 6). 10. 4 Vo^ - 3 y/¥x + 2 Vc^ + Vd2x - 2 V(6 + dfx. 11. 6 Vx + 3 V2x - 5 V3x - 2 V4x + Vl2x - \/l8x. 60 ALGEBRA TO QUADRATICS 80. Multiplication and division of radicals. For these pro- cesses we have the following EuLE. Follow the usual laivs of operation (§ 10), using also the assumption of § 77. Beduce each term of the result to its simplest form. " The operations of this section are limited to the case where the radicals are of the same index. Radicals of different indices as Vs and V^ must first be reduced to the same index. See § 87. EXERCISES 1. Multiply V2 - V3 by V2 - VS. Solution : V2 - V3 V2-V8 2 _ V6 - Vl6 + V2i = 2-V6-4 + 2V6 = - 2 + V6. 2. Divide ^_tZ_ by v^ + \/y. Vxy x-\-y y/xy ( Vx2 - Vxy + Vy2) ( Vx + y/y) * Solution : — — —— = — ^ ^ Vx + y/y y/xy ( Vx + Vy) _ Vx2 Vxy Vy2 Vx2/ Vxy V^ <^--</!- Carry out the indicated operations and simplify : 3. VlO . V5. 4. V^ . ■^. 5. V28 . V7. • 6. Vf ^ Vf- 7. (a -6x^1 8. VS-Vff. 9. ( V7 - V3) ( V3 - V2). 10. (- 1 + V3)^ 11. (6 V3 + V6) (6 V2 - 2). 12. y/lTc • VTOc. * Since as + 63= (a + ft)(a« - a6 + ft*), if o= V^, 6= Vy we have IRRATIONAL NUMBERS AND RADICALS 61 13. (a + 6-V^)(Va+V&). 14. (8 + 3 Vs) (2 - VB). 15. VVx + Vy • V Vx - s/y, 16. V6 + 2 V6 . V 6 - 2 V5. 17. Vx + Vjc2 - 1 . Vx - Vx2 - L 18. (x2 + y2) ^ (a; ^ _,. y ^), 19. --V^. 20. (4Va-V3^)(V^ + 2V3x). X 21. VS.Jp. \4a 25. (^.^Ij. 36.V^.Vg 27, ^*-^ + 66 y ' \x/ \ ox^ — to2 :fl^.(.t^-,%). 28. (5-l^).(l - 1). Vxy \y ^1 VVx Vy/ 29. (2 V6 - Vl2 - V2i + Vis) V2. 30. (3\/8 + Vl8+V60-2\/72)V2. 31. (5 v^ - 4 V32 + 3 V60 - 3 V54) V3. 32. ( V9^T6 + 3 A^)(V9x + 6 - 3 \^). 33. [(V7 4 V3 + Vl0)(V7 + V3-Vi0)7. 34. (2 V30 .. 3 V6 + 5 V3) ( V8 + V3 - Vs). 35. (2V^ + V8-Vl2)(^V30-fV3+V2). 37. Find the value of ^ V2i - V| + 2 Vs - V6- V3 4- V5 to three decimal places. 81. Rationalization. The process of rendering the irrational numerator (or denominator) of a fractional expression rational without altering the value of the fraction is called the rationaliza- tion of the numerator (or denominator) of the fraction. This is usually accomplished by multiplying both numerator and denominator of the given fraction by a properly chosen radical expression called the rationalizing factor. 62 ALGEBRA TO QUADRATICS The principles in accordance with which this rationalizing factor is selected are the following: Principle I. Since {a ■i-b)(a — b)=a^ — b% the rationalizing factor of -sfx ± Vy is sjx hF Vy. Principle II. Since {a^ — a'b-\-lF) (a + 5) = a' + h^, the ration- alizing factor of -sfx -\- Vy is V^ — V^ + Vy^ and conversely. Since (a^ -\- ah -\- If) (a — b)=a^ — b^, the rationalizing factor of -y/x — -Vy is V^ + Vxy + Vy^, and conversely. EXERCISES ^Nationalize the denominators of the following : - Va + Vx Va — Vx Solution : By Principle I the factor which will render the denominator rational is Va + Vx. Va + Vx Va + Vx Va + Vx a + x + 2 Vox Thus 2. Va—Vx Va—Vx Va + Vx 1 2+V2 + V3 Soltttioii : This problem requires a twofold rationalization. 1 (2+V2)-V3 2 + V2 + V3~ [(2+ V2) + V3][(2 +\^) - V3] ^ 2+-v^-V3 ^ 2+V^-V3 ~ (2 + V2y-S ~4 + 4V2 + 2-3 _ 2+V2-V3 _ (2 + v^-V3)(3-4\^ 3 + 4v^ ~ (3 + 4V2)(3-4v^) _ 6 + 3-v^-3V3-8\/2-8 + 4V6 "" 9 - 16 . 2 2 + 6v^+3V3-4V6 ^2-V3 Solution : ^2-2 + ^6 + ^^ Vi + v'e + v^ ^-.■^S ^-^3 ^2+^6+V3-« 2-3 = ^(^ + V6 + ^9). IRRATIONAL NUMBERS AND RADICALS 63 5 7-V5 g V3+V2 2+V3 3+V6 V3-V2 8 /- 3, V3+V2 V2-V4 V2+3Vi 10. ^/^«±^. 11. I^±^^. \a-y/x \a-Va2-l 12 2V6 ^2 1+3V2-2V3 * V2+V3+V6 ' V6_}.V3+V2 -- 2\/T6 -g Va + ic + Va — x V3 + V5 + 2 \/2 Va + x - Va - x 16 2 j^ V6-V5-V3 + V2 Va + 1 + Va- 1 V6 + V5 - V3 - V2 18. Show that ^^ ~ "^ = - .10 • ■ -. V2-fV3 19. Show that ^^^-^^ = 17.48 .... V8- -V7 20 V(l + a)(H-6)- -V(i -a)(l -&) 21. V(l + a)(l + 6) + V(l Show that ^^+^^ - -a)(l -h) V6_ = .168. V3 + V5 + V6 - V5 82. Solution of equations involving radicals. We prove the important Theorem. WTien an equation in x is multiplied hy an expres- sion in Xy the resulting equation has, in general, solutions which the first one did not possess. Let ^ = represent an equation contaming x which is satisfied by the values X = a, by ■ - n. Let 5 be an expression which vanishes when X = ay p, • ' V. Then the expression is satisfied not only when x = a, bj • - j n, but also when x = a,l3,-';v. . 64 ALGEBRA TO QUADRATICS Example. The equation x — 2 = has X = 2 for its only solution, while the equation , (X - 2) (X - 3) = has in addition the solution x = 3. If in the course of a problem it is necessary to multiply an equation by any expression involving the variable, the solutions of the resulting equation must be substituted in the first one to ascertain if any solutions have been introduced which did not satisfy the original equation. Solutions which have been intro- duced in the process of solving an equation, but which do not satisfy the original equation, are called extraneous solutions. It may be shown in a similar way that raising the equation in Xj A =B, to any power introduces extraneous solutions. EXERCISES 1. Solve Vaj + 19 + Vx + 10 = 9. Solution : '^ Vx + 19 = 9 - Vx + 10. X + 19 = 81 + X + 10- 18 V»Tl0. -72=- 18 Vx + 10. 4 = Vx + 10. 16 = x + 10. x = 6." Check : V6 + 19 + V6 + 10 =5 + 4 = 9. 2. Solve Va; + 19 - Vx + 10 = - 9. Vx + 19 = - (9 - VxTlO). Simplifying, we get x = 6. Check : V6 + 19 - V6 + 10 =+6-4 = 1 9^-9. Thus our result satisfies only the equation which was introduced in the course of solving the problem, and is extraneous. The Original equation has no solution. Solve and check, noting all extraneous solutions : 3. VSx -1=5. • 4. Vi« -8 = 2. 5. V2x + V3x = 1. 6. Vx + V3x = 2. 7. V5x-7=v'4x + 3. 8. 5Vx-7 = 3v^-l. IRRATIONAL NUMBERS AND RADICALS 65 9. Vx + 1 + Vx + 2 = 3. 11. 2\^-V2x = 2+v^. 13. V37 - 7 V6x^=^ = 4 10. Vl3 4-4v^^^ 6. 12. Vx + 4 + \^+T=L 14. Vx2 - 7 x + 19 = v^^^. 15. 7 + Va;2-lla; + 4 = jc. 17. a; — Vax (1 + x) + 1 - x = 1. 19. |(7V5 + 6)-5=|(3v^-l). I 4 16. 8 + V(x - 10) (X - 6) = X. 18. V2 (X + 1) + V2x + 15 = 13. 20. 2\/3 + 3V2x = 3V2 + 2V3a 21. 6 4- Vx 8 23. V7 X + 2 = 25. V2X-1 -Vx 6x + 6 V7X4-2' 2 (x - 3) V2x-10' 22. 24. 4. 5 + v^ 5-Vx a — Vto 26 3 Vox a + V6x 2 6 + 3 Vox 6x+ 10 26. V9 X + 10 = 27. VTT- VT 1 + VI -X 1 - VIT^ 29. X — ax : Vx = Vx : X. 30. 2v^^T2 + V^T2 = i^^±^. V8x + 8 1 + 2V3X-6 11 + 2 V3x- 5 V4X+9 28. Va - X + V6 - X = V6 31. 1 + 3V3X-6 11 + 5V3X-5 32. V9x + 7 + V4X + 1 = V25x + 14. 33. Vx + 15 + Vx - 24 - Vx - 13 = Vx. 34. Vx - 7 + Vx - 2 - Vx - 10 = VxT5. 35. (VS-7)(Vx-3) = (Vx-6)(Vx-5). 36. (a + \^) Vx:(6-Vx) Vx = a + 1:6-L 37. (4V5-7):(5V^-6) = (V^-7):(V^-6). 38. ( Va V6 — V6 Va) Vx = a V6 Vx — 6 VaVx. CHAPTEE VII THEORY OF INDICES 83. Negative exponents. We have already seen (§ 16) that a^-a"* = «" + "' (1) when n and m are positive integers. We now assume that this law still holds when one or both of the numbers m and n are negative or fractional. If we let a-m = , a'" then since the law (1) holds when n and m are any integers. This notation may be expressed verbally as follows : Principle. A factor of numerator or denominator of a frac- tion may be changed from the numerator to the denominator, or vice versa, if the sign of its exponent he changed. 84. Fractional exponents. Since (p. 57) Va • Va = a, it is natural to devise a notation for Va suggested by the law (1). If we let Va = a*, we have Va • Va = a* • a^ = a' "^ * = a^ = a. Furthermore, if we let -y^ — an it would be consistent with law (1) to write 1 11 11 s (a")'' = a« . a» = a" " = «"'. This notation we shall assume in general. Thus 66 THEORY OF INDICES 67 With the adoption of this notation we can attach a meaning :o any real number with any rational number for its exponent. This notation may be expressed verbally in the following Principle. The numerator of a fractional exponent indicates a power, the denominator a root. 85. Further assumptions. The operation of multiplication is subject to the following laws of exponents : I. Commutative law of rational exponents : (^a'^y = a""- = a'--'' =(a'-y. II. Associative law of rational exponents : (yj = a'?'- -^ = ««"•« = (a'^y. The laws of operation (§ 10) defined for integral values of the symbols we also assume when the symbols are expressions with rational exponents. 86. Theorem. a''lf= {ctby, where r is any rational number as _. We raise both sides of the equation a'b'' = (aby to the g-th power separately and show that the results are equal. Since r = -> l(abyY = [(abyj = {aby = aPbP. p p p p p p p p Also (or by = (a«55) = (a^^»«)(a«6«) • • • {a'^^) q terms p p P P P P q terms g terms =iJm' == a^hP, {ariry = l(abyj. ing the qih root and taking the principal root, we obtain a'b"- = (aby. • 68 ALGEBRA TO QUADRATICS EXERCISES yl. Express in simplest form with positive exponents : , , 36a-26-ic-5 (a) 9a26-2c-i Solution : By Principle, § 83, ^^^"^^~^^"^ 9a26-2c-i _46-i&+2c-ic+3 ~ a2-a2 ^46c ~ a* ' (b) (c) — - ^ ^ (xi2/l)-i ' 35x-2?/6z-4* 3a-i6-2 6a2x-i gft^c Va6v^ 4X-22/-4' 56-ic2' ^'^ v'^&-^aJ62c*' ,.. 5 11 / — I — \ — ; /., V 4x-«y-3 15a'263-m ^"•^ • ' 5 a- 4 6-"' 14x«2/«-3 • 2. Arrange in order of magnitude the following : (a) Vi, 71, 7|. Solution : We first ask. Is (f )i > (|)^ ? Raise both numbers to the sixth power. We obtain (|)3 and (f)2, or If and f, er 2if and 2|. Thus Vl>7l- Now compare (f)^ and (|)io Raise both numbers to the fourth power. We obtain . (|)2 and |, or 1| and If. Thus VI > Vl. THEORY OF INDICES 69 Now compare (§)' and (|)^. Raise both numbers to the twelfth power. We obtain (f)* and (|)3, or fl and \\^-, or 6yV and 6f|. Thus -s/i>\^h The order of magnitude is then Vli VI? Vf • (b) V^, VI, </h (c) 4^, V8, -Ws. (d) Vli, VI- (e) V3, ^5, Vl5. 3. Perform the indicated operations, (a) \/2 . V3. Solution : ^2 • V3 = 2' • 3^ = 2^ 3^ = (22 . 38)i = V'4^ = v^lOS. V3 V5 . ''V3.V2 87. Operations with radical polynomials. These operations follow the rules for the same operations previously given, pro- vided the assumptions and principles of §§ 83-86 are observed. EXERCISES 1. Divide x^ — y^ by ^/x — y/y. 2. Extract the square root of 4 x - 12 x^ 2/3 + 9 2/5 + 32 x^ - 48 yi + 64. 3x + 3x-i-6 3. Simplify xi- 3x2 + 3x-^-x-i 4. Divide J-Va^ -J-V^ by sJ-> a'2 52 5. Extract the square root of 1 2. 6. Multiply - 3x-5 + 2^ by ^ - ?^. X* x^ 6-1 7. Extract the square root of -ix2y-2- i^yx-i + ^y2x-2-^xy-^ + 25?. 49 2 Id 7 7 8. Multiply V^ - x» + x^ f- Vx^ -x+v^-lby Vx + 1. QUADRATICS AND BEYOND CHAPTEE VIII QUADRATIC EQUATIONS 88. Definition. An equation that contains the second bnt no higher power of the variable is called "a quadratic equation. The most general form of the quadratic equation in one variable is ax'^ + hx-{-c = 0, (1). where we shall always assume a, b, and c to represent rational numbers, and where a =^ 0. Every quadratic equation in x can be brought to this form by transposing and simplifying. 89. Solution of quadratic equations. The solution of a quad- ratic equation consists in finding its roots, that is, the numbers (or expressions involving the coefficients in case the coefficients are literal) which satisfy the equation. The common method of solving a quadratic equation consists in bringing the member of the equation that involves the variable into the form of a perfect square, i.e. into the form x^-\-2Ax + A\ For example, let us solve Transpose 8, x^-^2x = S. If now we add 1 to both sides of the equation, the left-hand member will be a perfect square, x^-\-2x-{-l = 9. 70 QUADRATIC EQUATIONS 71 Express as a square, (x + 1)^ = 9. Extract the square root, x -\-l =±S. Transpose, ic = — 4 or 2. Both — 4 and 2 satisfy the equation, as we see on substituting them for x. Thus (-4)^ + 2(-4)-8 = 0, and 2^ + 2 • 2 - 8 = 0. Consider now the general case. Let us solve ax'^ -{- bx -\- c = 0. Transpose c, ax^ -\- bx =— c. b c Divide by a, x^ -^ - x = / bV Add ( 77— I to both members to make the left-hand member a \2aJ perfect square, b b'' _ c b"" _ -4.ac-\-b^ ,a 4a^ a 4a^ 4 a^ 4 ac / by b^-4 Express as a square, ( x + — — I = — —^ Extract the square root. b V^^ — 4 ac x-\-w- = ± 2a 2a 2a Transpose, x = ^^ * (1) The roots are -b-\- V^^ - 4 ac _ -b - -s/b^ -4.ac Xi — „ f X2 — „ 2 a 2a That the equation can have no other roots appears from § 96. * This expression for the roots, _ / — X = » 2a may be used as a formula for the solution of a quadratic equation. Thus to solve the equation 2a;2-3a; — 6=0 we may substitute in the formula a = 2, 6 = - 3, c = — 6, and obtain 3±\^+48 3±V57 Thus Xi-. 4 4 3+V57 3-V57 72 QUADRATICS AND BEYOND One should verify the fact that both a-iid — satisfy (1) and are consequently roots of the 2a equation. They are, in geueral, distinct from each other. For particular values of the coefficients to be noted later (§ 98) the roots may be equal or complex (i.e. of form a -\- /3 V— 1, where a and yS are ordinary rational or irrational nimibers). We may sum up the process of solving a quadratic equation in the following EuLE. Write the expression in the form aa^ -\- hx -\- c = 0. Transpose the term not involving x to the right-hand side of the equation. Divide both sides of ^he equation by the coefficient of a^. Add to both members the square of one half of the coefficient of X, thus making the left-hand member a perfect square. Rewrite the equation, expressing the left-hand member as the square of a binomial and the right-hand member in its simplest form. Extract the square root of both members of the equation, not omitting the ± sign in the right-hand member. Transpose the constant term, leaving x alone on the left-hand side of the equation. The two values obtained on the right-hand side by taking the + and — signs separately are the roots sought. Check by substituting the solutions in the original equation, which should then reduce to an identity. 90. Pure quadratics. A quadratic equation in which the coeffi- cient of the term in x is zero is often called a pure quadratic. Its solution is found precisely as in the general case, excepting that we do not need to complete the square. Thus let us solve ax^ -f- c = 0. Transpose c, ax^ =— c. Divide by a, ic* = — -. QUADRATIC EQUATIONS 73 Extract the square root, ^ = ± _,' The roots are Xi = + -v/— -? x^ = ~ a/ EXERCISES Solve and check the following : 1. 3x2 -6x- 10 = 0. Solution : Transpose 10, 3 x^ — 6 x = 10. . Divide by 3, x2 - 2 x = J/- Add the square of \ the coefficient of x, i.e. 1, to both sides, x2_2x + l = -Lo + l = Y- Express as a square, (x — 1)2 = -LS-. Extract the square root, x — 1 = ± "V^- x = i±V¥. Check: 3(l±^'-6(l±^)-10 = 0. 3±6^| + |-6:F6Vir_io = 0. /13 2. 8x2 + 2x -3 = 0. Solution : Transpose 3, 8 x2 + 2 x = 3. Divide by 8, x2 + i x = f . Add the square of \ the coefficient of x, i.e. ^^j, to both sides, ^' + \^ + -h = \ + iz = il' Express as a square. (^ + \Y = f f- Extract the square root X + 1 = ± |. ^=±'J-\ = -lov\. Check: 8. (1)2 + 2. 1- 3 = 1 + 1-3 = 0. 8(-f)^ + 2(-f)- -3 = 11-1-3 = 1-1-3=1-1 3. x2-ax = 0. 4. x2 = 169. 5. x2-ix = i 6. fx2 = 560. 7. x2 + x-l = 0. 8. 19x2 = 5491. 9. 3x2-7x = 16. 10. x2 = .074529. 11. 3x2+ ll = 5x. 12. x2 - tl X = 1. 13. x2 + X - 66 = 0. 14. 20x2 + x = 12. 1 = 0. 74 QUADRATICS AND BEYOND 15. 7a;2 + 9x = 100. 16. 6x2 + 6x = 66. 17. 14x2 - 33 = 71x. 18. 5x2 + 13 = 14x. 19. x2 - 8x + 15 = 0. 20. 91 x2 - 2 X = 45. 21. x2 + 2 X - 63 = 0. 22. x2 - 6x + 16 = 0. 23. x2 - lOx + 32 = 0. 24. 6x2 + 26^ = 25 ix. 25. 6x2 - 13x + 6 = 0. 26. 15x2 + 527 = 178x. 27. 2x2 + 15.9 = 13.6 X. 28. (x - 1)2 = a(x2- 1). 29. a'^{b-x)^=h^{a-x)^. 30. 13x2-19 = 7x2 + 5. 31. ax2 - (a2 + l)x + a = 0. 32. (a - x)(x - 6) = - ab. 33. 14 x2 + 45. 5 X = - 36. 26. 34. a^ {a - x)2 = b'^{b- x)2. 35. (a-x)2+(x-6)2 = a2 + 62. 35. (a -x)(x-6) = (a -x)(c-x> 37.^' = ^. 38. 2x + l = 3. b d X 33 15x^810^ 40. x2 + ? = 50. 2 3x 7 - - 2 X 1050 yio« + a^,^ + a^ 5 41. — = 4<c. 1 = -. 3 7x 6 + xa + x2 43 ^ + ^^ :^ 2x + l 44 x3- 10x2 + 1 ^ ^ _ g ■x + 3 x+6* 'x2-6x + 9 .^ ax + b mx — n ac ^^ 1^ 4 45. = 46. = 0. 6x + anx — m xx — 2x-3 47. -— ^— = -. 48. — - — + = X - 1. 49. te2 — mx + n n 9 2 x — 3 x-2 _ 3 (8 - x) gQ (a-x)3 + (x-6)8 _ a^-b^ }x+ 14~ 28-x ' * " 51.'^— + ^-^ = 12. 52. 2x- 3 X- I 53 5 + X ' 8 - 3 X _ 2 X -^ ' 3 — X X x-2 55. (^'=8(^-15. 56.5^ \x -b) \x-bj 9 57. a2 - x2 = (a - X) (6 + c - x). (a- -X)- -(X- -6) a + &■ 16- ■ X 2(x -11) x-4 4 X • -6 12 2x- X — T* 3x- X - 4-1 -3 5x X -14 -4 5x- ^V 3x- -1_ ?4 ■X — 58. i= + ^ = ^. 59 2_^±^ + J?- = ^Ili + ^Z^. 18 x + 4 4 6 60. {X - a + 6) (X - a + c) = (a - 6)2 - x^. QUADKATIC EQUATIONS 75 91. Solution of quadratic equations by factoring. When the left-hand member of an equation can be factored readily, this is the most convenient method of solution. It also illustrates very clearly the meaning and property of the roots of the equation. Example. Solve x^ -}- 2 x — 15 = 0. Factor the left-hand member, (x + 5) (x — 3) = 0. The object in solving an equation is to find numbers that substituted for the variable satisfy the equation. But since zero multiplied by any number is zero (p. 3), any value of x which causes one factor of an expression to vanish makes the whole expression vanish. If in this case x = 3, our equa- tion in factored form becomes (3 + 5) (3 - 3) = (3 + 5) • = and is satisfied. If we let x = - 5, the other factor becomes zero, and the equation reduces to the identity (5_5)(6-3) = 0(5-3) = 0. Thus the numbers 3 and — 6 are solutions of the equation. 92. Solution of an equation by factoring. We have imme- diately the EuLE. Transpose all the terms to the left-hand member of the equation. Factor that member into linear factors. The values of the variable that make the factors vanish are roots of the equation. EXERCISES Solve and check the following : 1. 6x2 + x = 15. 3. 6x2 -X- 6 = 0. 5. 13x2-38x = 3. 7. x2-40x + 111 = 0. 9. x2-18x- 208 = 0. 11. x2-3ax-4a2 = 0. 13. (x2 _ 1) + (X - 1)2 = 0. 15. (2x - l)(x -I- 2) -f (X - 1) (X - 2).= - 4. 16. (7x - l)(x -f 3) - (4x - 3)(x - 1) =24. 2. 6x2-f-7x = 3. 4. 5x2-17x+6 = 0. 6. 2x2 -6x- 26 = 0. 8. 13x2 -40x + 3 = 0. 10. 3x2 -26x + 36 = 0. 12. (X - a)2 - (x - 6)2 = 0. 14. (3x_6)2_(9x-M)2 = 0. 76 QUADRATICS AND BEYOND 17. (3x + 1) (X + 1) - (4a; + 3) (x - 1) = - 2. 18. (X - a)(4ax - 6) + (x - 6)(4ax - 6) = 0. 19. 3x-4Vx^^ = 2(x + 2). Solution : Transpose, x — 4 = 4 Vx — 7. Square, x2 - 8 x + 16 = 16 x - 112. Transpose, x2 - 24 x -f 128 = 0. Factor, (x - 16) (x - 8) = 0. The roots are x = 16. x = 8. Check : 3-16-4 Vl6 - 7 - 2 (16 + 2) = 48 - 12 - 32 - 4 = 0. 3-8-4 VS~^ - 2(8 + 2) = 24 - 4 - 20 = 0. In the following examples, as always, the quadratic equation should be solved by factoring when possible. Recourse to the longer but sure method of completing the square is always available. When an equation is cleared of fractions or squared in the process of bringing it into quadratic form (1), §88, extraneous solutions may be introduced. The results should be verified in every case and extraneous solutions rejected. 20. a + Va2 - x^ = x. 21. VxT~5 = x - 1. 22. 2x-^V2x-l = x + 2. 23. 1- 6x + V5(x + 4) = 0. 24. Vll-x + \/x-2 = 3. 25. V2X + 1-2 V2x + 3 = 1. Hint. Square twice. 26. Va(x-6) + V6(x-a) = x. 27. VxTS + V2X-3 = 6. 28. 2 V3 + X - 4 VS^^ = V60. 29. Vl + ox - Vl - ax = X. 30. V5x- 1 - V8-2x = Vx^. 31. Vx + 7-V5x-2 = 3. 32. x - 10 = |(x - 1)-V2x-1. 33. Va2"^^ + V&2 + X = a + &. 34. V4 - x + V6 - x = V9 - 2; X X \b 35. (a2-62)(x2+l) = 2(a2 + 62)x. 36. x + 2 a V2(a2 + 62) - x = 3a2 + 62. 37. xV^:r2 + 2V^T2zzV^H^. 38. -J^-^ + \f-^ = 2- gg \^a-\-x-\-Va-x _a .^ Va + Vx _ 2 Vx (x+a)2 Va + X - Va - X X Va-Vx y/a + Vx a{x-a) Hint. Rationalize. Va — X 4- Vx — 6 ja — x ^^ ^ , y/b-Va_ 1 Va—Vb 41. V'J^ + v^ ^ /«E^. 42.^+; Va - X - Vx-6 \ X - 6 V6 Vx Va QUADRATIC EQUATIONS 77 ^2 2a-(l + a2)a; ^ 26 + (l + 62)x 4 44. 2v^+4-3V2x-3 Vx + 4 45. (a - a;)2 + (& _ x)2 = f (a - cc) (6 - x). 46. a&x2 - (a + 6) (a6 + 1) x + (a6 + 1)2 = 0. 47. V2x-2+V3x + 7 = V2x+ll + V3x-8. 48. V{a + x) (X + 6) + V(a - x) (x - b) = 2 Vox. 49. 2 72 a + 6 + 2 X - Vl0a + 6-6x = Vl0a + 96-6x. 93. Quadratic form. Any equation is in quadratic form if it may be written as a trinomial consisting of a constant term and two terms involving the variable (or an expression which may be considered as the variable), the exponent in one term being twice that in the other. By the constant term is meant the term not containing the variable. Thus X -S V ^ + 13 = 0, a ;-^ + x"^ - 3 = , a2x-2» _ (a + 6) a;-« + 62 ^ 0, x2 — 2x — S— Va;2 _ 2 x — 3 + 17 =^ are all in quadratic form. In the last the whole expression x2 _ 2 x — 3 is taken as the variable. It is usually convenient to replace by a single letter the lower power of the variable or expression with respect to which the equation is in quadratic form. EXERCISES Solve and check the following : 1. X - 8 v^ + 15 = 0. (1) Solution : Let y/x = y. Then x = y2. Substituting, (1) becomes y2 _ 3^ _j. 15 = 0. Factor, (y - 6) (y - 3) = 0. The roots are y = 5, y = S. Thus Vx = 6, Vx = 3, or X = 25, X = 9. Check: 25 -8-5 + 15 = 0; 9-8-3 + 15 = 0. 78 QUADRATICS AND BEYOND 2. x-i - 5x-§ + 4 = = 0. Solution : Let X-3 = y. Then X-t = y1. Substituting, (1) becomes 2/2_ 5y + 4 = 0. Factor, (y - 4) (y - 1) = 0. The roots are 2/ = 4, y = \. Since x-l-^- ' . x3 ^' we have ' VX2 ^2 Thus 4' X2 = 1 1 = 64' "^-=^8' and 1' X2 = :1; x=±Vl = ±l. (1) 3. 2Vx2-2x-3 + x2 - 2x = Solution : Add — 3 to both members and rearrange terms, x2 _ 2x - 3 + 2 Vx2 - 2x - 3 - 3 = 0. (1) Let Vx2 - 2 X - 3 = y. Then x2-2x -3 = 2/2. Substituting, (1) becomes ^2 ^ 2 y — 3 = 0. . Factor, (y + 3) (y - 1) = 0. The roots are y = —3, y = 1. Hence y2 _ ^2 _ 2 x - 3 = 9, or x2_2x + l = 13. Extract the square root, x — 1 = ± "s/l3. The roots are x = 1 ± VlS. Also x2 - 2 X - 3 = 1, or x2-2x + l = 5. Extract the square root, x — 1 = ± V5. The roots are x = 1 ± VB. 4. x8 - 1 = 0. 5. x8 - 8 = 0. 6. x8-l = 0. 7. x5 + 8xi = 9x. 8. X - 6 x-i = 1. Hint. Divide by Vx. This factor cor- responds to the root z = 0. 9- 4x3 + 5x^-1 = 0. 10. 10x4-21 = x2. QUADRATIC EQUATIONS 79 11. 2x^-3x^-\-x = 0. 12. 3v^ + 6-l^=4. 13. ax^p + bxp + c = 0. 14. 3 ic* - 7 x2 - 6 = 0. 15. 4x6 - 14x3 + 6 = 0. 16. x* - 13x2 + 36 = 0. 17. x~ 12Vx + ll = 0. 18. x3 + 4x^= 16ixVx2. 19. ■\/x3-2Vx + x = 0. 20. 8X-6 + 999x-3= 125. 21. 7^x6 + 5xv^ = 66. 22. 2(V5 - 3)=* - 3 = v^. 23. (x2 - 10) (x2 - 3) = 78. 24. (x2 _ 1)2 + (a;2 + i) = 2. 25. ( v^ - 1)^ + v^ = ^z. 26. x^ + x^ = (28 + 2-8)x V^xl 27. (Vx- 3)(v'x - 4) = 12. 28. (X + a)^ + (x + 6)* = (a - b)K 29. (2x2-3)2-(x2 + 4)2 = 7. 30. 2x2 + 3 Vx2 - x + 1 = 2x + 3. 31. (2x + 3)' + (2x + 3)-^:^0. 32. 8(8x - 5)8 + 5(5 - 8x)« = 85. 33. x4-4(a+6)x2+16(a-6)2 = 0. 34. 4x2 + 12xVTTi = 27(1 + x). 35. (2x2-3x+l)2=22x2-33x + l. 36. (x2_ 5x+7)2 -(x-2)(x-3) = l. 40. ^^ = x-8. 37. x2 + 5 = 8x + 2 Vx2 - 8x + 40. 38. 2 V(x - 4)2 + 4(x - 4)"^ = 9. x-4 2 + x 42. a = x2 + 6 + -. 44.1+^- , ^' , =3. 2 2(l + vT+^)' .- 7 , 41 Vx 97 . 5 ._ 2x (x-l)2 . 45. X* H = h x5. 46. h ^^ = 4. X ^^ (x-l)2 2x 47. (X - 1)2 + 4 (X - 1) - 5 = 0. Hint. I.ety = -^^: then- = ^^~^^'' ' ^ ' ^ ' " (X — 1)2 y 2x 48. (x2 + 2)^+ ^ =4x2 + 8. Vx2 + 2 94. Problems solvable by quadratic equations. The principle of translating the problem into algebraic symbols, explained in § 55, should be observed here. The result should be verified in every case. It may happen that the problem implies restrictions that are not expressed in the equation to which the problem leads. In this case some of the solutions of the equation may not be consistent with the problem ; for instance, when the 80 QUADKATICS AND BEYOND variable stands for a nuinber of men fractional solutions should be rejected. If only such results are obtained, the problem is seK-contradictory. Often negative solutions should be rejected, as when the result indicates a negative number of digits in a number. Imaginary or complex (p. 72) results in general mean that the conditions of the problem cannot be realized. PROBLEMS 1. The product of ^ and ^ of a certain number is 500. "What is the number ? 2. There are two numbers one of which is less than 100 by as much as the other exceeds it. Their product is 9831. What are the numbers ? 3. The sum of the square roots of two numbers is VtI. One of the numbers is less than 37 by as much as the other exceeds it. What are the numbers ? 4. A man sells oranges for -^^ as many cents apiece as he has oranges. He sells out for |3. How many had he ? 5. If the perimeter of a square is 100 feet, how long is its diagonal ? 6. A man sells goods and makes as much per cent as ^ the number of dollars in the buying price. He made $246. What was the buying price ? 7. Two bodies A and B move on the sides of a right angle. A is now 123 feet from the vertex and is moving away from it at the rate of 239 feet per second. B is 239 feet from the vertex and moves toward it at a rate of 123 feet per second. At what time (past or future) are they 850 feet apart ? 8. What is the number twice whose square exceeds itself by 190 ? 9. What numbers have a sum equal to 63 and a product equal to 612 ? 10. The sum of the squares of two numbers whose difference is 12 is found to be 1130. What are the numbers ? 11. By what number must one increase each factor of 24 • 20 so that the product shall be 540 greater ? 12. What numbers have a quotient 4 and a product 900 ? 13. Two numbers are in the ratio of 4 : 5. Increase each by 15 and the difference of their squares is 999. What are the numbers ? 14. If 4^ is divided by a certain number, the same result is obtained as if the number had been subtracted from 4^. What is the number? 15. Separate 900 into two parts such that the sum of their reciprocal values is the reciprocal of 221. QUADRATIC EQUATIONS 81 16. The denominator of a fraction is greater by 4 than the numerator. Decrease the numerator by 3 and increase the denominator by the same, and the resulting fraction is half as great as the original one. What is the original fraction ? 17. The numerator and denominator of a fraction are together equal to 100. Increase the numerator by 18 and decrease the denominator by 16, and the fraction is doubled. What is the fraction ? 18. A number consists of two digits whose sum is 10. Reverse the order of the digits and multiply the resulting number by the origingll one, and the result is 2944. What is the number? 19. The sum of two numbers is 200. The square root of one increased by the other is 44. What are the numbers ? 20. The difference of two numbers is 10, and the difference of their cubes is 20630. What are the numbers ? 21. Around a rectangular flower bed which is 3 yards by 4 yards there extends a border of turf which is everywhere of equal breadth and whose area is ten times the area of the bed. How wide is it ? 22. Two bicyclists travel toward each other, starting at the same time from places 51 miles apart. One goes at the rate of 9 miles an hour. The number of miles per hour gone by the other is greater by 5 than the number of hours before they meet. How far does each travel before they meet ? 23. A printed page has 15 lines more than the average number of letters per line. If the number of lines is increased by 15, the number of letters per line must be decreased by 10 in order that the amount of matter on the two pages may be the same. How many letters are there on the page ? 24. A merchant buys goods for a certain sum. The cost of handling them was 5% of their cost price. He sells for §504, gaining as much per cent as T^Q the cost price was in dollars. What was the cost price ? 25. A man had $8000 at interest. He increased his capital by $100 at the end of each year, apart from his interest. At the beginning of the third year he had $8982. 80. What per cent interest did his money draw ? 26. Two men A and B can dig a trench in 20 days. It would take A 9 days longer to dig it alone than it would B. How long would it take B alone ? 27. A cistern is emptied by two pipes in 6 hours. How long would it take each pipe to do the work if the first can do it in 5 hours less time than the second ? 28. A party procures lunch at a restaurant for $16. If there had been 5 less in the party, each member would have paid 15 cents more without affecting the amount of the entire bill. How many were in the party ? 82 QUADRATICS AND BEYOND 29. A party pays $12 for accommodations. Had there been 4 more in the party, and if each person had paid 25 cents less, the bill would have been $16. How many were in the party ? 30. A grocer sells his stock of butter for $16. If he had had 6 pounds less in stock, he would have been obliged to charge 10 cents more a pound to realize the same amount. How many pounds had he in stock ? 31. A man buys lemons for $2. If he had received for that money 50 more lemons, they would have cost him 2 cents less apiece. What was the price of each lemon ? 32. It took a number of men as many days to dig a ditch as there were men. If there had been 6 more men, the work would have been done in 8 days. How many men were there ? 95. Theorems regarding quadratic equations. In this and the following sections we prove several theorems concerning quad- ratic equations. Similar theorems are later proved in general for equations of higher degree. Theorem. If a is a root of the equation ac(^-\-hx-\-c = 0, (1) then X — a is a factor of its left-hand member , and conversely. The fact that a is a root of the equation is equivalent to the assertion of the truth of the identity aa^ -\- ha ■\- G =i Oj by definition of the root of an equation (§ 53). Divide ax^ 4- Jx -f c by ic — a as follows : X — a\ ax^ -\- bx -\- c \ax + (^ 4- ao^) (b + aa) X -{- c (b -f- aa) X — aa^ — ba aa^ -{- ba -\- c Since the remainder vanishes by hypothesis, ax^ + bx -{- c is exactly divisible hy x — a. Conversely, we have already seen (p. 3) that ii x — a is a, factor of an equation, a is a root, since replacing x hy a would make that factor vanish. QUADRATIC EQUATIONS 83 EXERCISES Form equations of which the following are roots. 1. 2, 6. Solution : Since 2 and 6 are roots, (x — 2) and (x — 6) are factors, and the equation having these as factors is (X - 2) (X - 6) = x2 - 8 X + 12 = 0. 2. 1, - 1. 3.-3,-4. 4. V2, - V2. 5. - 2, - 6. 6. - 4 V2, V32. 7. V27, - 3 VS. 8. 2 + V3, 2 - V3. 9. a + V6, a - V6. 96. Theorem. A quadratic equation has only two roots. Given the equation ax^ + ?)X + c = with the roots a and /3, to prove that the equation has no other root, as 7, distinct from a and /3. Since a and j8 are roots of the equation, x — a and x — /3 are factors. Thus our equation may be written in the form a(x-a:)(x-/3) = 0. (1) If now 7 is a root, it must satisfy (1), i.e. a(7-a)(7-/3) = 0. But in order that any product of numerical factors should vanish one of the factors must vanish. Tlius either a = 0, or 7 — or = 0, or 7-/3 = 0. But, by hypothesis, 7 5^ <x and 7 ^ /S, so the last two factors cannot vanish. Thus a = 0. This would, however, reduce our equation to a linear equa- tion, which is contrary to our hypothesis that the equation is quadratic. Thus the assumption that we have a third distinct root leads to a contradiction. CoKOLLARv I. If a quadratic equation is satisfied by more than two distinct values of the variable, then each of the coefficients vanishes identically. The above proof shows that the coefficient of x^ must vanish. In the same way it can be shown that b = c = 0. Corollary II. If two quadratic expressions have the same value for more than two values of the variable, then their coefficients are identical. Let ax2 + 6x + c = a'x^ + b'x + c' for more than two values of x. Transpose, and we obtain (a - a')x^ + (6 - b')x + c - c' = 0. We have then a quadratic equation satisfied by more than two values of X. Thus by Corollary I each of its coefficients must vanish. Thus a' = a, 6' = 6, c' = c. 84 QUADRATICS AND BEYOND This theorem taken with § 95 is equivalent to the statement that a quad- ratic equation can be factored in one and only one way. Thus if ax"^ + 6x 4- c = a (x — or) (x — /3), we cannot find other numbers 7 and 5 distinct from a and /3 such that ax2 + 6x + c = a (x — 7) (x — 5), for then the equation would have roots distinct from a and /3. 97. Theorem. If the equation x2 + 6x + c = 0, (1) where b and c are integers, has rational roots, those roots must be integers. P For suppose - to be a rational fraction reduced to its lowest terms and a root of (1). Then ^, + ^-^ + = 0. q2 q or p2 + ipq + c^2 = 0, which gives p'^ = - q{bp-{- cq). Thus some factor of q must be contained in p (§ 69), which contradicts the hypothesis that the fraction - is already reduced to its lowest terms. 98. Nature of the roots of a quadratic equation. The equation ax^ -^ bx -\- c = (1) has as roots (§ 89) x^ = "^ ^ ^^, (2) _ ^ -l_ V^'^ - - 4c ac 2a -b-^b^- - 4: ao 2a ■ (^) These expressions afford an immediate arithmetic means of determining the nature of the roots of the given equation when a J b, and c have, numerical values and a =^ 0. In fact an inspec- tion of the value of 6^ — 4 ac is sufficient to determine the nature of the roots of (1). 'mM I. When h^ — Jf.ac is negative, the roots are iikhg%naty (§89). II. When y^ — Ji^ac^Oy the roots are real and equal. In this b case x^=x^=-—- III. When y^ — Ji^ac is positive, the roots are real and distinct. IV. When ¥ — Jiac is positive and a perfect square, the roots are real, distinct, and rational. QUADRATIC EQUATIONS 85 In case IV, if J^ — 4 ac = A, _ — h + Va _ — ^, - Va ^' ~ 2 a ' ^"^ " 2 a ' The converses of these four cases are also true. For instance, if the roots of (1) are imaginary, from (2) and (3) it is clear that P — 4: ac must be negative. The expression A = Z>^ — 4 ac is called the discriminant of the equation ax^ -\- bx -^ c = 0. EXERCISES 1. Determine the nature of the roots of the following equations without solving. (a) 3x2 _4x-l =0. Solution : A = (- 4)2 - 4 . 3 . (- 1) = 16 + 12 = 28 and is then positive. Thus by III the roots are real and distinct. (b) 3x2 - 7x + 6 = 0. (c) 6x2 - X - 1 = 0. (d) 3x2 + 4x + l = 0. (e) x2-4x + l = 0. (f) 2x2 -6x-9=:0. (g) 2x2-4x-2 = 0. (h) 4x2 + 12x + 9 = 0. (i) 2x2 + 6x - 4 = 0. (j) 4x2- 28x + 49 = 0. (k) 4x2 + 12x + 5 = 0. 2. Determine real values of k so that the roots of the following equations may be equal. Check the result. (a) (2 + A;)x2 + 2A:x + l = 0. Solution : Here 2 -\- k = a, 2fc = 6, l = c. Thus A = 62-4ac = 4fc2_4.(fc + 2).l = 4 ^2 _ 4 A: _ 8. Since the roots of an equation are equal when and only when its dis- criminant equals zero (§ 98, II), the required values of k make A = and are the roots of 4^2 _ 4^ _ 8 = 0, or fc2 - A: - 2 = 0. Solve by factoring, k^-k-2=:{k-2){k + l) = 0. Thus the values of k are /c = 2, k = — 1. Check : Substituting in the original equation for /c = 2, we get 4x2 + 4x + l = (2x + l)2, and for A: = — 1 we get x2 — 2 x + 1 = (x — 1)2. 86 QUADRATICS AND BEYOND (b) x2 + fcx + 16 = 0. (c) a;2 + 2a; + A;2 = 0. (d) x2-2A:x + l = 0. (e) 3A;x2-4x-2 = 0. (f) A;x2-3x + 4 = 0. (g) x^ -\- 4kx -\- k^ + 1 = 0. (h) A:2x2 4- 3x - 2 = 0. (i) (A:2 + 3) x2 + fee - 4 = 0. (j) 3/cx2 + A-x - 1 = 0. (k) x2 + (3A; + l)x + 1 = 0. (1) x2 + 3x + A; - 1 = 0. (m) x2 + 9A;x + 6fc + ^ = 0. (n) 4 A;2x2 + 4 A-x - 125 = 0. (o) 2 x2 - 4x - 2 A; + 3 == 0. (p) (A: + l)x2 + A:x + A; + 2 = 0. (q) A;x2 + (4 A; + l)x + 4 A: - 3 = 0. (r) 2(A:4- l)x2 + 3A:x + A;-1 = 0. (s) (A: - l)x2 + 5A:x + 6A; + 4 = 0. (t) (2A; + 3)x2-7A;x+^^^ = 0. (u) (A;-l)x2+(2A; +l)x+ A: + 3 = 0. CHAPTEE IX GRAPHICAL REPRESENTATION 99. Representation of points on a line. Let us select on the indefinite straight line AB sl certain point as a point of refer- ence. Let us also select a certain line, the length of which for the purpose in hand shall represent unity. Let us further agree that positive numbers shall be represented on ^J5 by points to the right of 0, whose distances from are measured by the given -3 -2 -1 +1 +2+3 1 numbers, and negative numbers similarly by points to the left. Then there are certainly on AB points which represent such num- bers as 2, — 3, ^, — 1^, or, in fact, any rational numbers. Since we can divide a line into any desired number of equal parts, we are able to find by geometrical construction the point correspond- ing to any rational number. Furthermore, by the principle that the square of the hypotenuse of a right triangle equals the sum of the squares of the other two sides, we can find the point corresponding to any irrational number expressed by square-root signs over rational numbers. More complicated irrational num- bers cannot, however, in general be constructed by means of ruler and compasses, but we assutne that to every real number there corresponds a point on the line, and conversely , we assert that to every point on the line corresponds a real number. This assump- tion of a one-to-one correspondence between points and real num- bers is the basis of the graphical representation of algebraic equations. This amounts to nothing more than the assertion that every real number, rational or irrational, as, for instance, — 6, 2 + V3, V3, tt, represents a certain distance from on AB, and conversely, that whatever point on the line we may select, the distance from to that point may he expressed by a real number. . 87 88 QUADRATICS AND BEYOND '<Y X axis T^ 100. Cartesian coordinates. We have seen, that when the single letter x takes on real values all these values may be repre- sented by points on a straight line. When, however, we have two variables, as X and y, which we wish to represent simultaneously, we make use of the plane. Just as we determined arbitra- rily, on the line along which the single variable was represented, an arbitrary point for the point of reference and an arbitrary length for the unit distance, so now we select an indefinite line along which x shall be repre- sented, and another perpendicular to it along which 7/ shall be represented. The former we call the X axis ; the latter the Y axis. The intersection of these axes we take as the point of reference for each. This point is called the origin. We select a unit of distance for x and a unit of distance for 1/ -v^hich may or may not be the same, according to the problem under discussion. As before, we represent positive numbers on the X axis to the right, and negative numbers to the left. Positive values of y are represented above the X axis, and negative values below it. The arrowhead on the axes indicates the positive direc- tion. Any pair of values of x and y, written (x, ?/), may now be represented by a point on the plane which is x units from the Y axis and y units from the X axis. Thus if x = 0, y = 0, written (0, 0), the point repre- sented is the origin. The point (3, 0), i.e. a; = 3, ?/ = 0, is found by going three units of x to the right, i.e. in the positive direction of x and no units up. The point (4, 3), i.e. a = 4, 2/ = 3, is found by going four units of x to the right and three units of y up. The point (— 3, 1) is found by 3,4) YA 0(0,7) U',3. 3.0) {4, -2) GRAPHICAL REPRESENTATION 89 going three units of x to the left and one unit of y up. The point (— 3, — 4) is found by going three units of x to the left and four units of y down. In fact, if we let both x and y take on every possible pair of real values, we have a point of our plane corre- sponding to each pair of values of (x, y). Conversely, to every point of the plane correspond a pair of values of (x, y). These values are called the coordinates of the point. The value of a?, i.e. the distance of the point from the Y axis, is called its abscissa; the value of y, i.e. the distance of the point from the X axis, is called its ordinate. If the point (x, y) is conceived as a moving point, and if no restriction is placed . on the value of the coordinates so that they ^ ., T 'PIT ^>^<^ Quadrant ist Quadrant take on every possible pair oi real values, (-,+; (-|-,+) ith Quadrant every point in the plane is reached by the ^ moving point (X, y). 3rd Quadrant The X and Y axes divide the plane into (-'-) four parts called quadrants, which are num- bered as in the figure. The proper signs of the coordinates of points in each of the quadrants are also indicated. EXERCISES The following exercises should be carefully worked on plotting paper, which can be bought ruled for the purpose. 1. Plot the points (2, 3), (0, 4), (- 4, 0), (- 9, - 2), (2, - 4). 2. Plot with the aid of compasses the points (l, V'2), (V3, — V2), (2+V3, 2-V3),(-V2, -V2). 3. Plot the square three of whose vertices are at (— 1, — 1), (— 1,+ 1), (+ 1, — 1). What are the coordinates of the fourth vertex? 4. Plot the triangle whose vertices are (2, 1), (—6, — 2), (—4, 4). 5. Plot the two equilateral triangles two of whose vertices are (6, 1), (—6, 1). Find coordinates of the remaining vertices. 6. If the values of the coordinates (x, y) of a moving point are restricted so that both are positive and not equal to zero, where is the point still free to move ? 7. If the coordinates {x, y) of a moving point are restricted so that con- tinually y = 0, where is the point still free to move ? 90 QUADRATICS AND BEYOND 8. What is the abscissa of any point on the T axis ? 9. The coordinates of a variable point are restricted so that its ordinate is always 2. Where may the point move ? 10. If both ordinate and abscissa of a point vanish, can the point move ? Where will it be ? 11. Plot the quadrilateral whose vertices are (0, 0), (— 6, — 3), (5, — 5), (—1, — 8). What kind of a quadrilateral is it? 12. The coordinates of three vertices of a parallelogram are (— 1, — 1), (6, 2), (—1, — 6). Find the coordinates of the fourth vertex. 13. The coordinates of two adjacent vertices of a square are ( — 1, — 2) and (1, — 2). Find the coordinates of the remaining vertices (two solutions). Plot the figures. 14. The coordinates of two adjacent vertices of a rectangle are (— 1, — 2), (1, — 2). What restriction is imposed on the coordinates of remaining vertices ? 15. The coordinates of the extremities of the bases of an isosceles triangle are (1, 6), (1, — 2). Where may the vertex lie? What restriction is imposed on the coordinates (x, y) of the vertex ? 101. The graph of an equation. The equation a? = 2 ?/ is satis- fied by numberless pairs of values (x, ?/); for example, (2, 1), (0, 0), (1, ^), (— 2, — 1) all satisfy the equation. There are, how- ever, numberless pairs of values which do not satisfy the equation ; for example, (1, 2), (2, - 1), (- 1, 1), (0, — 1). The pairs of values which satisfy the equation may be taken as the coordinates of points in a plane. The totality of such points would thus in a sense represent the equation, for it would serve to dis- tinguish the points whose coordi- nates do satisfy the equation from those whose coordinates do not. After finding a few pairs of values which satisfy the above equa- tion we note that any point whose abscissa is twice the ordinate, i.e. for which cc = 2 ?/, is a point whose coordinates satisfy the equation. Any such point lies on the straight line through the origin and the point (2, 1). We can then say that those points I GRAPHICAL REPRESENTATION 91 and only those which are on the straight line represented in the figure have coordinates which satisfy the equation. This line is the graphical representation or graph of the equation. The equation of a line or a curve is satisfied by the coordinates of every point on that line or curve. Any point whose coordinates satisfy an equation is on the graph of the equation. 102. Restriction to coordinates. Iri § 100 it was seen that a moving point whose coordinates were unrestricted took on every position in the plane. We now see that when the coordinates of a point are restricted so as to satisfy a certain equation (as x = 2 3/), the motion of the point is no longer free, but restrained to move along a certain path. Thus, for instance, the equation a; = 4 means that the path of the moving point is so restricted that its abscissa is always 4. Its ordinate is still unrestricted and may have any value. This shows that the plot of a? = 4 is a straight line four units to the right of the Y axis and parallel to it, for the abscissa of every point on that line is 4, and every point whose abscissa is 4 lies on that line. EXERCISES Determine on what line the moving point is restricted to move by the following equations. Draw the graph. 1. x = 6. 2. x = 0. 3. ?/ = f. 4. X = y. h. y = 2. 6. 3 X = y. 7. 2x = y. 8. y = 0. 9. x = -3. 10. X = 32/. 11. 3y = - X. 12. X + y = 0. 13. 6x = ll. 14. 2/=-3. 15. 2x = -3?/. 16. x = -2y. 17. x = -l. 18. 2x-62/ = 0. 103. Plotting equations. Plotting an equation consists in find- ing the line or curve the coordinates of whose points satisfy the equation. Thus the process of § 101 was nothing else than plot- ting the equation x = 2y. This may be done in some cases by observing what restriction the equation imposes on the coordinates of the moving point ; but more often we are obliged to form a 92 QUADRATICS AND BEYOND table of various solutions of the equation, and to form a curve by joining the points corresponding to these solutions. This gives us merely an approximate figure of the exact graph which becomes more accurate as we find the coordinates of points closer to each other on the line or curve. EuLE. Wlien y is alone on one side of the equation, set x equal to convenient integers and compute the corresponding values of y. Arrange the results in tabular form. Take corresponding values of x and y as coordinates and plot the various points. Join adjacent points, making the entire plot a smooth curve. When X is alone on one side of the equation integral values of y may be assumed and the corresponding values of x computed. Care should be taken to join the points in the proper order so that the resulting curve pictures the variation of y when x increases continuously through the values assumed for it. By adjacent points we mean points corresponding to adjacent values of x. Any scale of units along the X and Y axes that is convenient may be adopted. The scales should be so chosen that the portion of the curve that shows considerable curvature may be displayed in its relation to the axes and the origin. When there is any question regarding the position of the curve between two integral values of x, an intermediate fractional value of x may be substituted, the corresponding value of y found, and thus an additional point obtained to fix the position of the curve in the vicinity in question. EXERCISES Plot: 1. x2 -4a; + 3 = y- Solution : In this equation if we set x = 0, 1, 2, 3, etc., we get 3, 0, — 1, 0, etc., as cor- responding values of y. Thus the points (0, 3), (1, 0), (2, - 1), (3, 0), etc., are on the curve. These points are joined in order by a smooth curve. X y X y 3 -1 8 1 -2 15 2 -1 3 4 3 6 8 6 16 \ W^ 1/ 1 i \ / \ / \ / ' /■ i \ / \ f \ / *x — — — - GRAPHICAL REPRESENTATION 93 2. y = x^-lx + l. 4. y = x^ - Sx + 2. 6. y = x^ — 2x + 1. 8. 2/ = 2x2-6x + 7. 10. y = 2x^-6x-3. 3. y = x2 + 1. 5. y = x^ — 4x. 7. y = x'^ + 6x + 5, 9. y = 2x2^-3x + 4. 11. 2/ = x2-12x + ll. 104. Plotting equations after solution. When neither x nor y is abeacly alone on one side of the equation, the equation should be solved for y (or x) and the rule of the previous section applied. It should be noted that when a root is extracted two values of y may correspond to a single value of x. EXERCISES Plot: 1. 2x2+ 3y2 = 9. Solution 32/2 = 9 2/2 = 3 2x2, 3/ = ± V3-fx2. Assuming the various integral values for x, we obtain the following table and plot: X y ±V3 = ±1.7 1 ±VI = ±i-5 2 ±Vi=± -67 + 3 imaginary X y -1 -2 -3 ±VJ = ± -57 imaginary N: In this example, when x is greater than 3 or less than — S, y is imaginary. Thus none of the curves is found outside a strip x = ±S. To find exactly where the curve crosses the X axis, the equation may be solved for x and the value of x corresponding to y = found. Thus X = ± VF^I^'- If 2/ = 0, X = ± Vl = 2-1. This point is included in the plot 2. X2/ = 4. Solution : 4 94 QUADRATICS AND BEYOND Form table for integral values of x. i X y -1 -4 -2 -2 -3 -f -4 -1 -6 -1 -8 -\ -12 -i Since this table does not give us any idea of the curves between -\- 1 and — 1, v^e supplement the table by assuming fractional values for x. 3. x^ = y\ 5. xy = — 1. 7. x2 + ?/2 = 16. 9. a;2 + 2/2 = 25. 11. 2x2/ + 3x = 2. 13. xy + 2/2 = 10. 10 — 2/2 Hint, a; = y 4. X2/ = 16. 6. X2/ = X + 1. 8. x2 - 2/2 = 9. 10. x2 + X = 12 2/. 12. x2 + 9 2/2 = 36. 14. X — 2/ + 2 xy = 0. 2/ Hint. x = — -• 1 + 22/ 2/ X 6 -1 8 -1 12 -i y -8 12 15. 6x2+2x + 32/2 = 0. 16. x2 + 2x + 1 = y2 _ 3^. 105. Graph of the linear equation. The intimate relation between the simplest equations and the simplest curves is given in the following theorems. Theorem I. The graph of the equation y = ax is the straight line through the origin and the point (1, a), where a is any real number. The proof falls into two parts. First. Any point on the line through the origin and the point (1, a) has coordinates that satisfy the equation. Let P (Figure 1) with coordinates (x', y') be on the line OA. By similar triangles V or ax' GRAPHICAL KEPRESENTATION 95 Thus the coordinates of any point on the line satisfy the equation. C«'y) Figure 1 Figure 2 Second. Any point whose coordinates satisfy the equation lies on the line. Let the coordinates (x\ y') of the point P (Figure 2) satisfy the equation. Then we have or y = ax', ^, = a. x' Let the ordinate y' cut the line at B. Then by the first part of the proof BC = ax', EC or x' a. Thus Hence P lies on the line. v' EC a=—, = — TJ or ti' = EC. x' x' -^ Theorem II. The graph of any linear equation in two vari- ables is a straight line. The general linear equation Ax-{-By-{-C = (1) may be written in the form y = ax-\-b, (2) A C where a = — — and b= — —f provided ^ ^ (§ 7). It E = 0, B E the equation Ax -\- C = may be put in the form C 96 QUADRATICS AND BEYOND provided A =^ 0. This is evidently the equation of a straight line parallel to the Y axis (§ 102). li B = and A = 0,-we have no term left involving the variable. Thus the only case for which the theorem demands proof is when B ^ 0, and the equation may be reduced to form (2). By Theorem I we know that the graph oi y = ax is a straight line. If, then, we add to every ordinate y of the line y = ax the constant h, the locus of the extremi- ties of the lengthened ordinates will lie in a straight line, as one can easily prove by Geometry. But any point (cc, y) on the upper line is such that its ordinate y is equal to the ordinate of the lower line, i.e. ax, and in addition the constant h ; that is, y = ax -\- h. Also, since the upper line is the locus of the extremities of the lengthened ordinates, every point whose coordinates satisfy the equation y = ax + h i^ on this upper line. Thus the equation (1) has a straight line as its graph. CoEOLLARY. Two lines vjhose equations are in the form y = ax-\-h, (3) yz=ax-\-V (4) are 'parallel. ^ For the value of the ordinates of (3) corresponding to a given abscissa, say x^^ is obtained from the ordinate of (4) corresponding to the same abscissa by adding the constant h — h\ Thus each point on (3) is always found h — V units above (below if h — b' is negative) a point of (4). Thus the lines are parallel. 106. Method of plotting a line from its equation. Since the equation y = ax-\-h\^ satisfied by the values (0, h), the graph cuts the Y axis h units above (below if b is negative) the origin. Since it is satisfied by the values (1, a -\-b), the graph passes through the point reached by going one unit of x to the right of (0, b) and a units up (down if a is negative). These two points determine the line. We may then plot a linear equation by the following GRAPHICAL REPRESENTATION 97 Rule. Reduce the given equation to the form y — ax-\-h. Plot the point (0, h) as one of the two points that determine the line. From this point go one unit of x to the right and a units of y up {down if a is negative) to find a second point that lies on the line. Draw the line through these two points. EXERCISES Plot: 1. 6x + 2y-5 = 0. Write in the form y = ax + h, and we have y=_3x + f. a =-3; 6 = f. Plot the point (0, |). From this point go one unit of x to the right and three units of y down to find the second point, which helps determine the line. 2. 6i 32/ + ll = 0. 11 Yk J rV / / io. n / / / / f i / ' 3. x-y = 0. 5. X + y = 4. 7. 2 X - y = 4. 9. Sx-y = 0. 11. x-8y = 16. 13. x = 8(2-2/). 15. X - y - 1 = 0. 17. x + 2/ + l = 0. 19. 12x-3y = l. 4. x — y — 5 = 0. 6. 2x = 6(l-y). 8. 12x + 10y = 5. 10. 16x-10y = 4. 12. 2x + y + 3 = 0. 14. 2x-6y-l = 0. 16. 2x-2y-6 = 0. 18. 3x-6y-4 = 0. 20. 7x-8y-9 = 0. 107. Solution of linear equations, and the intersection of their graphs. The process of solving a pair of independent linear equations consists in finding a pair of numbers (x, y) which satisfy them both. Though each equation alone is satisfied by- countless pairs of values (x, y), we have seen that there is only one pair that satisfy both equations. Since a pair of values which 98 QUADRATICS AND BF.YOND satisfy an equation are the coordinates of a point on its graph, it appears that the pair of values that satisfy simultaneously two equations are the coordinates of the point common to the graphs of the two equations, that is, the coordinates of the points of intersection of the two lines. EXERCISES Find the solutions of the following equations algebraically. Verify the results by plotting and noting the coor- dinates of the point of intersection. 1 3^- ^' 3x- -iy + 16: = 0, -y - -7 = Solution 3x- '4.y + 16 = 3x- y - 7 = Sy -23 = . y = ¥• Substituting in (2), 3x = 7 + ¥ X = (1) (2) ¥• To plot (1) and (2) we get the equa- tions in the form y = ax + h and apply the rule. Thus 2/ = fx + 4. 2/ = 3x-7. S'^^'iM /\l ^ \ i^'f)'! y "{0.4)' i ^^ /~ z / y - i I "7^ t X . 7 r K'.-i) '-t . JCo.r) 2 2x + 3y = 6, 7x + y = 2. 5. 3x-2y = l, 3x + 2?/ = 5. g 2x + y = 3, • 8x-72/ = l. 11 x + y = -i, ' 4x-3y = 5. 14. 17. X + y = 5, 4x-2y = 28. x-y = -4, 2x + 6y = 16. 6. 9. 12. 15. 18. X + y = 5, 4 x-3y = -7, 3x + y = l. 4x + y = ll. 3x-7y = 9, 7. X + 2/ = - 7, X + 2 y = 3. 2x-3y = 6. 2x-5y = 0, x-y = S. 10. X + 2 y = - 10, 2x-y = 0. x-y = 1, 13. x-y = l, 2x-8y = 3. 2x-42/ = -16. X - 2/ = 2, 16. 6x-5y = 5, 4x- 52/ = 9. 2 X + 3 y = — 20. 3x + 2|/ = 9, 8x-2/ = 2. 19. 2x + 6y = -20, • 3x + y = 2. GRAPHICAL REPRESENTATION 99 108. Graphs of dependent equations. We have defined (§ 57) dependent equations as those that are reducible to the same form on multiplying or dividing by a constant. Thus two dependent equations are reducible to the same equation of the form y = ax + b. Hence dependent equations have as their graphs the same straight line. We see now the geometrical meaning of the statement that dependent equations have countless common solutions. Since their graphs have not one but countless points in common, being the same line, it is clear that the coordinates of these countless points will satisfy both equations. 109. Incompatible equations. By our definition (§ 60) incompatible equa- tions have no common solution. Since every pair of distinct lines have a common point unless they are parallel, we can foresee the Theorem. Incompatible equations have parallel lines as graphs. Let the equations ax -{- by -{- c = 0, (1) a'x + b'y + c' = (2) be incompatible. This is true (§ 60) when and only when ab' - a'b = 0. (3) (4) (5) This may be done if neither 6 nor ¥ equals zero. If both b and b' vanish, the lines (1) and (2) are both parallel to the Y axis and hence to each other, which was to be proved. But if only one of them vanishes, say & = 0, then by (3) a = (§ 5), in which case (1) does not include either variable. Thus we may assume that neither b nor 6' vanishes and that (4) and (5) may be obtained from (1) and (2). By (3) a _a/ b~V Our equations (4) and (5) become a c a c' y = --x_-, which represent parallel lines, by the Corollary, p. 96. This theorem completes the discussion of the graphical representation of the possible classes (§ 61) of pairs of linear equations. Let us then assume (3). Write (1) and (2) in the form y = a c —b'-b y = 100 QUADRATICS AND BEYOND EXERCISES Plot and solve : 1. 8x + 2y = 3, 4x + y = 8. 3. 10x-5y = 2 X - y = 3. 15, 5. x-1y = l, 4x-2Sy = 56. 7. x-Sy = 2, 6x-lSy = 36. 2x + 6y = 1, x-^Sy = 1. 2x-Sy = 6, 8x- 12y = 24. 12x-6y=18, 2x-y=l. 2x-S + y = 0, 4x -7 -\-2y =0. 110. Graph of the quadratic equation. Let y = ax^ -{- bx -\- c, (1) where as usual a, b, and c represent integers and a is positive. If we let X take on various values, y will have corresponding values and we may plot the equation as in § 103. A root of the quadratic equation ax^-}-bx + c = (2) is a number which substituted for x satisfies the equation, that is, gives the value y = in (1). Thus the points on the graph of (1) which represent the roots of the equation (2) are the points for which y = 0, that is, where the curve crosses the X axis. The numerical value of the roots is the measure of the distance along the X axis from the origin to the points where the curve cuts the axis. Since this distance is always a real distance, only real roots are represented in this manner. Theorem. If the graph of (1) has no point in common with the X axiSf the equation (2) has imaginary roots, and conversely. Every equation of form (2) has two roots either real or imagi- nary (§ 89). If the graph of (1) has no point in common with the X axis, there is no real value of x for which y = 0, i.e. no real root of (2). The roots must then be imaginary. Conversely, if (2) has only imaginary roots, there is no real value of X which satisfies it, i.e. which makes y = in (1). Thus the curve has no point in common with the X axis. GRAPHICAL REPRESENTATION 1,01 This suggests the following universal Principle. Non-intersection of graphs corresponds to imagi- nary or infinite-valued solutions of equations. 111. Form of the graph of a quadratic equation. Consider the equation ^ o rr ^ ^ y ^2x^+1 x-\-2. ^ (1) By substituting for x a very large positive or negative number, say X =± 100, y is large positively. Thus for values of x far to the right or left the curve lies far above the X axis. If we assign to t/ a certain value, say 3/ = 2, we can find the correspond- ing values of x by solving a quadratic equation. Thus in (1) let or . 2x^ + 1 x = 0. The roots are x^=— 3^, x^ = 0. Hence the points (— 3^, 2) and (0, 2) are on the curve (§ 101). That is, if we go up two units on the Y axis, the curve is to be found three and one half units to the left and also again on the Y axis. If in (1) we let y = — 4, the corresponding values of x are very nearly equal to each other (— 1^ and — 2), which means that the curve meets a line parallel to the X axis and four units below it at points very near together. The question now arises, Where is the bottom of the loop of the curve ? This lowest point of the loop has as its value of y that number to which correspond equal values of x. Hence we must determine for what value of y the equation (1), that is, the equation 2x^-{-lx + {2-y)=0, has equal roots. Comparing with the equation ax"^ + Jx + c = or Z - = a, 7 = --b,2- -y=-c. Thus the condition h 2-4ac = becomes 49-4 •2(2- -2/)=0, y = 49- 8 16_ 33 " 8 4i. 102 QUADRA'EICS AND BEYOND I as the corre-' Substituting this value of y in (1), we get spending value of x. This gives a single value of y for which the values of x are equal; hence the graph of (1) is a single fes- toon as in the figure. If we take the gen- eral equation ax^ -{- bx -{- c = y, we find precisely similarly that the bottom of the loop is at a point whose ordinate is P — Aac A y = - y X 2 -4 -.3+ or -3.2 + or - ^ - li or - 2 -If 4a 4a Thus we see again that if the discriminant is negative the graph is entirely above the X axis and both roots are imaginary (§§ 98, 110), since the ordinate of the lowest point of the loop is positive. If the discriminant is positive, the graph cuts the X axis and both roots are real. The results of this section enable us to determine a value of y from the coefficients which determine the lowest point of the loop of the curve precisely, and hence to show beyond question from the graph whether the equation ^as real or imaginary roots. EXERCISES Plot the following equations and determine by measurement the roots in case they are real. Find in each case the lowest point on the loop. 1. x2 + X + 1 = y. 4. x2 + 7 aj + 6 = y. 7. x2-6x + l = y. 10. x2 + 2x-l = 2/. 13. 2x2-x-3 = y. 3. x2-6x + 10 = ?/. 6. 3x2-7x-6 = y. 9. 2x2-9x + 7 = y. 2. x2 - 4 X + 7 = y. 5. x2-6x + 9 = y. 8. x2-6x + 5 = y. 11. x2 -4x + 4 = y. 14. 3 x2 + 8 X + 5 = y. 16. What is the characteristic feature of the plot of an equation whose roots are equal ? 12. 3: 4x-3 = y. 15. 4x2+12x+9 = y. * GRAPHICAL REPRESENTATION 103 112. The special quadratic adc^ + 6ic = O. When in the quad- ratic equation ax^ -\-hx^ G = 0, (1) c = 0, we can always factor the equation into ax^ -\- bx = X (ax -j- b) = 0, (—:)=»■ or Thus the roots are £Ci = 0, X2= Conversely, if a; '= is a root, then (§ 95) ic — 0, or x, is a factor and the equation can have no constant term. This affords the Theorem. A quadratic equation has a root eqtcal to zero when and only when the constant term vanishes. We show in a similar manner that both roots of the equation (1) are zero when and only when i = c = 0. EXERCISES 1. Prove the theorem just given by considering the expressions for the roots in terms of the coefficients (§ 89). 2. For what real values of k do the following equations have one root equal to zero ? (a) x2 + 6x - A: + 1 = 0. (b) 2x2 - 3x + A;2 - 1 = 0. (c) x2 + 6a; ^. ^2 ^. 1 ^ 0. (d) 2x2 - 4x + fc2 _ 3^ ^ 0. (e) 2x2 + 2fcx-2A;2-4A;-2=0. (f) 6x2 - 4x + 2 A;2 + fc + 7 = 0. 3. What is the characteristic feature of the plot of an equation which has one root equal to zero ? 4. For what real value of A: will both roots of the following equations vanish ? (a) - + 3x - 1 = 0. (b) x2 + (fc2 + 3)x + fc - 3 = 0. (c) x2 + (fc2 + l)x + l = 0. (d) x2+(fc-3)x + 2fc2-5fc-3 = 0. (e) x2 + (A: + l)x + /c2 -1 = 0. (f) (fc-3)x2 + (fc2_9)x + A;2-4A; + 3=a 104 QUADRATICS AND BEYOND 113. The special quadratic ax^ + c = O. This equation may be written in the form x^ -\- - = and factored * immediately into (^-*-^R)(^-^R)=o> which shows that the roots are equal numerically but have oppo« site signs. The roots are Xi =\/-^' ^2 xR Since in the equation ax^ -\- c = y the variable x occurs only in the term a:^, we get the same value of y for positive and negative values of x. Hence the loop which forms the graph of the equa- tion is symmetrical with respect to the Y axis. 114. Degeneration of the quadratic equation. The equation ax^ -\-hx -\- c =■ has the roots _ J 4_ V52 - 4 ac Xi — 2a Xo = We wish to find the effect on the roots x^ and x^ when a becomes very small. If we let a approach 0, then x^ approaches an expression of the form -? which must always be avoided. Rationalize the numerators and we get 4 ac 2 c 2a{- b - ■\/b^ - 4 ac) -b-\^b^-4: ac 4 ac 2 c 2a{-b + -^b^-4:ac) - b -{- ^b^ -4.ac As a approaches 0, evidently h^ — ^ac approaches J^, Xi ap- proaches — j> and x^j since its denominator becomes very small, • When — Is positive this involves real factors. If — is negative the factors are a a imaginary (§ 152). GRAPHICAL REPRESENTATION 105 increases without limit, that is, approaches infinity. Thus the quadratic equation approaches a linear equation when a approaches 0, and one of its roots disappears since it has increased in value beyond any finite limit. The loop-shaped graph of the quadratic equation must then approach a straight line as a limit when a approaches 0. This is made clear from the following figure, where a has the successive values 1, \, y^^, ^V, 0. In the figure the curves represent the following equations : x-'-l-2 = y.. (I) x" X 5-2-2 = ^- (H) 10-2-2 = 2'- (III) X^ X ^ 50-2-2 = ^- (IV) -f-2- (V) %%-A "■ Ml II 171 ^^^J 7 1 tsX^ t J~ >^ \^v ^ t t h- ^§^^A l f - ^ ^SSv^ t ^ y^^^ ^^> ^ ^'^,^ < N^^ ^^^>^ ^^^-L*- ^-.K In a similar manner we can show that when in the equation Jx + c = 0, & approaches as a limit, the root of the linear equa- tion becomes infinite. EXERCISES 1. "What real values must k approach as a limit in order that one root of each of the following equations may become infinite ? (a) Arx^ + Gx + l = 0. (c) (A;x - 1)2 - (X + 2)2 = (A: + a:)2. (e) V2A:x-H-\/6F=^=V^TT. X (b) (^2 + 1)3524. a; + 1 = 0. (d) A;2 + 4fc2a;2_(aj_i)2 4.2 (g) 1 fc + 1 X 1 X + 1 A;2 0. (f ) Vx-fc + Vx + A; = yfkx + 1. 1~\ fc2 / fcx-1 'a:x + 1 (i) (^ 1)^ ^ (fc (A; + 1) k (X + 1)2 (j) (A:2 - l)x2 + (A: - l)x + A:2 + 4A; - 5 106 QUADRATICS AND BEYOND 2. What real values must k and m approach as a limit in order that both roots of the following may become infinite ? (a) A:x2 + mx + 1 = 0. (b) (2fc-m)x2 + A:x-2 = 0. (c) 2 tec2 + (3m - 1 + h)x = 8x2 - 1. (d) ijc - l)x2 + (A; + m + l)x + 3 = 0. (e) x2 - X - 2(fc + m)a; = (fc + m) (x2 - 1). (f) (fc + m)x2 + 2 (fc + m) + 1 = x2 - 2 x. (g) (fc + m + l)x2 + (2 A: - m - l)x + 1 = 0. (h) (2A; + m + 2)x2 + (4A: + 2m + 3)x + 3 = 0. 115. Sum and difference of roots. Let x^ and x^ be the roots of x'' + bx + G = 0. (1) Then (§ 95) x — x^ and x — x^ are factors, and their product ^'^ — (^1 + ^2)^ + ^1^2 is exactly the left-hand member of (1). Consequently the equation x^ -\- bx -\- c = x^ — (xi -{- Xz) X + X1X2 is true for all values of x. Hence by § 96 - (^1 4- X,) = b, (2) X1X2 = c. (3) We may state these facts in the Theorem. The coefficient of x in the equation s(^-^hx-\-c=0* is equal to the sum of its roots with their signs changed. The constant term is equal to the product of the roots. EXERCISES 1. Prove the statement just made from the expression for the roots in terms of the coefficients (§ 89). 2. Form the equations whose roots are the following : (a) 6, 1. (b) i, I (c) I, 3. (d) - i - 6. (e) h h (f) -h+h (g) 2 + V3, 2 - V3. (h) - V3, V3. * We should for the present exclude the case where 6^ - 4c<0, since the roots ar, and a?j are then imaginary and we have not as yet dettned what we mean by the sum or the product of imaginary numbers. We shall see later that the theorem is also true in this case. GRAPHICAL REPRESENTATION 107 3. If 4 is one root of ic^ — 3 x + c = 0, what value must c have ? Solution : Let Xi be the remaining root. Then by (1) _ (xj + 4) = - 3, or xi = — 1. By (2) c = Xi.4=:(-l)4 = -4. 4. Find the value of the literal coefficients in the following equations. (a) x^-\-bx-9 = 0. One root is 3. (b) 0:2 + 4 X + c = 0. One root is 2. (c) ax2 + 3x-4 = 0. One root is 2. (d) ax2 + 3x + 4 = 0. One root is 7. (e) ax2 + 2 X + 6 = 0. One root is 6. (f ) x2 + 6x + 4 = 0. One root is - 1. (g) x2 - 6x - 6 = 0. One root is - 3. (h) x2 + 6x + 6 = 0. One root is - 6. (i) 2 x2 — 6 X — c = 0. One root is — 4. (j) x2 — 6 X + c = 0. One root is double the other. (k) x2 + c = 0. The difference between the roots is 8. (1) x2 — 5 X + c = 0. One root exceeds the other by 3. (m) x2 — 7 X + c = 0. The difference between the roots is 6. (n) x2 — C X + c = 0. The difference between the roots is 4. (o) x2 — 3 X + c = 0. The difference between the roots is 2. (p) x2 — 2 X + c = 0. The difference between the roots is 8. 116. Variation in sign of a quadratic. It is often necessary to know the sign of the expression ax2 + 6x + c for certain real values of x, and to determine the limits between which x may vary while the expression preserves the same sign. We assume as usual that a is positive. Theorem L* If the discriminant of ax^ + hx + cis positive, the quadratic is negative for all values of x between the values of the roots of the equation. For other values of x {excepting the roots) the quadratic is positive. * If a were negative, Theorem I would read as follows : If the discriminant is posi- tive, the quadratic is positive for all values of x between the values of the roots of the equation. For other values of x {excepting the roots) the quadratic is negative. When a is negative Theorems II and III may be modified in an analogous manner. 108 QUADKATICS AND BEYOND \ In § 98 we found that when the discriminant of a quadratic equation is positive the equation has two real roots. If two roots are real, the loop of the graph of the equation ax^ + bx + cz=y cuts the X axis in two points (§ 110) as in the figure. The roots are represented by A and B, and any real value of X between the roots is repre- sented by a point P in the line J.JB. Since the curve is below the X axis at any such point, the value of y, i.e. of the expression ax^ + &x + c for values of X between the roots, is negative. The value of the expression for any value of X greater or less than both roots is seen to be positive, since for such points, for example Q and R, the graph is above the X axis. Theorem II. If the discriminant of ax^ -\- hz + c is negative, the expression positive for all real values of x. When the discriminant is negative the entire graph of ax^ + &x + c = y is above the X axis (§ 111), and consequently for any real value of x the corre- sponding value of y, i.e. the value of ax^ + &x + c, is positive. Theorem III. If the discriminant of ax^ + bx + c is zero, the value of the expression is positive for all values of x except the roots of the equation ax^ -\- bx + c = 0. Hint. See example 16, p. 102. We may restate these three theorems and prove them algebraically as follows : Theorem IV. If the discriminant of the quadratic ax^ + bx-\- cis positive, the values of the quadratic and a differ in sign for all values ofx lying between the roots, and agree for other values. If the discriminant is zero or negative, the value of the quadraiic always agrees with a in sign. Case I. Since the discriminant is positive, the equation ax^+bx+c=(i has two unequal real roots, as Xi and x^, of which we will assume Xi is the greater, and we may write the quadratic in the form ax2 + 6x + c = a (x — Xi) (x — X2). Now for any value of x between Xi and X2 the factor x — Xi is negative, while X — Xa is positive, which shows that the quadratic is opposite in sign to a for such values of x. For other values of x both these factors are either positive or negative, and for such values the quadratic is of the same sign as a. GKArillCAL KEPRESENTATION 109 Case II. Since the discriminant 6^ _ 4 ^c is negative and the roots are * *!, * & ± V&2 - 4 ac .^ ^^ , -: . of the form — — — — - , we may write the quadratic 2a „ , , , / & V62 - 4 ac V , 6 . V62 - 4 ac\ V 2a 2a /\ 2a 2a / Now for any value of x the expression lx-\ ) is positive, and since - \ 2 a/ 62 — 4ac is negative, 4ac — 6^ is positive; and we observe that tlie last member of the equation has the same sign as a. Case III. Since the discriminant is zero, the roots are equal and the expression has the form ax2 4- &x + c = a (x — Xi)2, which has evidently the same sign as a, for any value of x. EXERCISES 1. Between what values of x is the expression Vx^ — 5 x + 4 imaginary ? Solution : The roots of x^ — 5 x + 4 = are 4 and 1. The discriminant A = 62 _ 4 ^c = 25 — 16 = 9 is positive. Thus by Theorem I or IV, if 1 < x < 4 f the expression under the radical sign is negative and the whole expression is imaginary. 2. For what values of k are the roots of (A:+ 3)x2 + A:x + l = (1) (a) real and unequal ? (b) imaginary ? Solution : a = k + S, b = k, c = 1. A = b^-4ac = k^- 4{k + S) = k'^-ik- 12. (a) If A > 0, the roots of (1) are real and unequal. The roots of A;^ - 4 A: - 12 are A: = - 2 and 6 Then, by Theorem II, if A; < - 2 or fc > 6, A > 0. (b) By Theorem I, if - 2 < A; < 6, A < 0, and the roots of (1) are imaginary. * See 5 162. t Road " 1 is le6S than x wliich is lew than 4 " or " a; is between 1 and 4." 110 QUADRATICS AND BEYOND 3. Determine for what values of x the following expressions are negative, (a) aj2 + 2x - 1/ (b) x^-Sx-h 4. (c) x2 - 11 oj + 10. (d) x^-15x-{- 60. (e)-x2-2x + l. {i) - x^ -\- 7 X + SO. 4. Determine for what values of k the roots of the following equations are (a) real and unequal, (b) imaginary. (a) 3 A;x2 - 4 X - 2 = 0. (b) aj2 + 4 fcx + A:2 + 1 = 0. (c) x^-{-9kx + 6k + l = 0. (d) x2 + {3k + l)x + 1 = 0. (e) (A;2 + 3)x2 + A;x - 4 = 0. (f) 2 x2 - 4x - 2 A; + 3 = 0. (g) fcx2 + (4A; + l)x + 4 A: - 3 = 0. (h) {k - l)x2 + 6kx -{- 6k + i = 0. (i) {k - l)x2 4- {2k+l)x + A: + 3 = 0. CHAPTEE X SIMULTANEOUS QUADRATIC EQUATIONS IN TWO VARIABLES 117. Solution of simultaneous quadratics. A single equation in two variables, as x^ -\- y^ = 5, is satisfied by many pairs of values, as (1, 2), (Vl> V|)> (2, 1), and so on, though there are at the same time numberless pairs of values that do not satisfy it, as (0, 1), (1, 1), (2, 3). Thus the condition that (x, y) satisfy a single quadratic equation imposes a considerable restriction on the values that x and y may assume. If we further restrict the value of the pair of numbers (cc, y) so that they also satisfy a second equation, the number of solutions is still further limited. The problem of solving two simultaneous equations consists in finding the pairs of numbers that satisfy them both. 118. Solution by substitution. In this method of solution the restriction imposed on (x, y) by one equation is imposed on the variables in the other equation by substitution. Example. Solve 2x^-\-y'^-l, (1) x-y = \. (2) Solution : Equation (2) states that x = 1 + y. Thus our desired solution is such a pair of numbers that (1) is satisfied and at the same time x is equal to y + 1. If we substitute in (1) 1 + y for x, we are imposing on its solution the restriction implied by (2). Thus 2(l + 2/)2 + 2/2=,i^ or \ 3 2/2 4- 4 y + 1 = 0. The roots are y = — 1, y = — i. Corresponding to ?/ = — 1 we get from (2) x = 0. Corresponding to y =— \ we get from (2) x = |. Thus the solutions are (0, — 1) and (|, — \). Ill 112 QUADRATICS AND BEYOND EXERCISES Solve the following : xy = 4. 2. x-y = 6, xy = 36. ■ x2 + 2/2 = bxy. xy — x = 0. 5. x-hy = a, x2 + 2/2 =, 5. x2 4-2/2 = 50, • 9x + 7 2/ = 70. '• x-Sy = 0. 8. 2x-3y = 4, x2 - y2 = 0. 9 xy = 12, 2x + 3?/ = 18. 10 x:y = 9:4, ^"- a;:12 = 12:y. 11. X2:y2=:a2:62^ a — X = b — y. 12 5x2 + 2/ = 3xy 2x-2/ = 0. *"• 2x-32/ = 0. 14 (a^ + y)C z-2y) = T, 3. -g 3x2-4y = 6x- 3x+42/ = 10. 2y% 16 x2 + ,= x:y = 2 2/2 + X - 18, :3. ^^ ax-by = cy, ^g x2 + 2 X2/ + 2/2 = 7 (x - y), a2x2 — 62y2 _ acx2/ + to2. ' 2 X — 2/ = 5. ^g ax2 + (a-6)x2/-&2/=^=c2, ^q 2x2-5x2/+2/2+10x+12y=100, ■ (x + 2/) : (X - 2/) = a : 6. 2j 7(x + 5)2 -9(2/ + 4)2 = 118, * X - 2/ = 1. x2 + 2/2 = 130, 23. X + y ^ x-y 2X-2/ + 1 8 25. x-22/ + l~3' x2 - 3 X2/ + 2/2 = 5. 27. X2/ + 72 = 6(2x + 2/), X 2 V 8 4X+2/-1 4X+2/-12 29. 2X+2/-1 2X+2/-12 3x + y = 13. 31. 10 9' x + 2 + 2/-l = '' 2 4 = 2f, 2x-32/ = l. 22. x2 + 2/2 = a2, X m 2/ n 24. x2 + 2/ + 1 3 2/2 + X + 1 2 x — y = l. 26. 1 + X + X2^3 1 + 2/ + 2/^ a; + 2/ = 6. • 28. 2/ + 6 X X — 2/ = 4. 30. «(x-2/)-52/ = 6 x-2/ 32. ? + ? = 3, x 2/ ^^Tri = ;- 5(2/-l) = 2(x + l). SIMULTANEOUS QUADRATIC EQUATIONS 113 119. Number of solutions. We have proved (p. 42) that two linear equations have in general one and only one solution. Theorem. A quadratic equation and a linear equation ham in general two and only two sohUions. If the linear equation is solved for one variable, say x, and this is substituted in the quadratic equation, we get a quadratic equa- tion to determine all possible values of the other variable (i.e. y), which must in general be two in number (§ 98). To each one of these values of y will correspond one and only one value of x, thus affording two solutions of the pair of equations. EXERCISES 1. When may, as a special case, a quadratic and a linear equation have only one solution? 2. When may a quadratic and a linear equation have imaginary solutions ? 3. rind the real values of k for which the following equations have (1) only one solution, (2) imaginary solutions. x2 + y2 = 16, (1) ^^^ x-y = k, (2) Solution : x = y + k. Substitute in (1), {y + fc)2 -{- y"^ = 16. or 2 y2 4- 2 A:y + A;2 - 16 = 0. As in § 98, a = 2; b = 2k; c = k^ -16. Hence A = b'^ - Aac = ^k^ - Sk^ + 12S = - Ak^ + 128. • (1) A = when 4k^ = 128, or k = ± 4 -^2. There is then only one solution. (2) A < when k^ > 32. The solution is then imaginary. ^^ 2x-y = k. x^-y^ = 9, ^^^ x-2y = k. 2x2 + 3y2 = 6, ^J^ x-ky = \. x + ky = b, ^""^ (C2 + 2/2 = 5. ^ y-x = k, ^ ^ x^y = k. x2 + 2/2 = 25, (') 4x-3y = k. 114 QUADRATICS AND BEYOND 1 . 120. Solution when neither equation is linear. In the exam- ples previously given one equation has been linear and the other quadratic in one or both variables. Often when neither of the original equations is linear a pair of equivalent (p. 41) equations one or both of which are linear may be found. These latter equa- tions may be solved by substitution. EXERCISES Solve the following equations. When neither equation is linear, we can often obtain by addition an equation from which by the extraction of the square root a linear equation may be found. 1. x2 + 2/2 = 17, (1) xy = 4. (2) Solution: x^ + y^ = 17 (3) Multiply (2) by 2, 2xy = 8 (4) Add x2 + 2 icy + 2/2 ^ 25 Extract the square root, a; + y = ± 5. (5) Subtract (4) from (3), x2 - 2 xy + 2/2 = 9. Extract the square root, x — y =±S. (6) Solve (5) and (6) as simultaneous equations, X + y = ±6, x~y = ±S. x = +i, +1, -1, -4. 2/ = +l, +4, -4, -1. Thus the solutions are four in number, (4, 1), (1, 4), (—1, — 4), (— 4, —1). The following exercise affords another case where a linear equation may be found by addition and extraction of the square root. 2. x^-hxy = 6, (1) xy + y^ = 10. (2) Solution : Add (1) and (2), x^-^2xy + y^ = 16. Extract the square root, « + 2/ = ± 4. Substitute in (1), x2 + x ( ± 4 - x) = 6, x2±4x-x2=:6, X = ± f = ± 1^. Substitute in (4), 2/ = 2^, — 2|. Thus our solutions are (- |, - 2|), (f, + 2J). SIMULTANEOUS QUADRATIC EQUATIONS 115 When neither of the original equations is quadratic, we can often find by division an equivalent pair of equations one of which is linear and the other quadratic, as in the following exercise. 3. x3 + 2/8=12, (1) X + 2/ = 2. (2) Solution: Divide (1) by (2), x2 - ajy + y2 ==: 6. (3) Square (2), x^ + 2 xy + y^ = 4 Subtract, — 3 xy = 2 xy=-l (4) Solve (4) with (2) by substitution. When the sum of the exponents of the variables is the same in every term, the equation is called homogeneous. Thus, z'^ + xy^O, 2x^y -3zy^ - 'iz^ - 3y^ = 0. When one equation is homogeneous and the other either linear or quadratic we may solve them as follows : 4. 6x2-7xy + 2y2 = o, (1) x2 - y = 4. (2) Solution : Divide (1) by y^, Let- = 2,« 622_7z + 2 = 0. y Solve for z, z = | or |. Thus ?=:ior? = ?. y 2 y 3 Solve (2) with 2 x = y and 3 x = 2 y. When both equations are homogeneous except for a constant term we may solve as follows : 5. x2-x2/ + 2y2 = 4, (1) 2x2-3x!/-2?/2 = 6. (2) Solution : Eliminate the constant term by multiplying (1) by 3 and (2) by 2, 3x2-3x2/+ 6?/2 = i2, (3) 4 x2 - 6 xy - 4 y2 = 12 (4) Subtract (3) from (4), x2 - 3 xy - 10 y2 = o * "We observe that y ^0. For if y = were a value that satisfies equation (1), x = would correspond. But (0, 0) does not satisfy (2); thus y = Ois not a value that can occur in the solutions of the equations. 116 QUADRATICS AND BEYOND X Divide by y^ and let - = 2, where y ^d, y 22 _ 32 _ 10 = 0. (S)' Factor, (z - 6) (2; + 2) = 0. The roots are _ = 6, - = - 2. (6) y y Solve (6) with (1). "When one equation is quadratic in a binomial expression we may solve as follows : 6. x-y -Vx-2/ = 2, (1) x^-ys = 2044. (2) Solution: Let Vx - ?/ = z. Then (1) becomes z2 - z = 2. Solving for z. z = 2 or - 1. Thus or Solve (3) with (2) as in exercise x-2/ = n (3) x-y = lj ^ ' 3. 7. x^ + y^ = xy = X -\- y. 8. X3 + 2/3 =7x2/ = 28(x + 2/). 2/3 + x^y = 4. x2^ = a, ^"- X2/2 = b. - x{y-l) = 10, ' y{x-l) = 12. 12 a^2 + 2/2-a, ' xy = b. ^^ x^y + xy^ = a, x^y — xy^ - b. .. x + X2/=:35, • 2/ + X2/ = 32. -g x{x^-{-y^) = 7, y{x^-\-y^) = l. 2x2-32/2 = 6, • 3x2-2 2/2 = 19. 3x2-22/2 = 16. -J. 5x2 + 2^2 = 22, 3x2-52/2 = 7. IQ ic2 + ccy + 2/2 = 2, • x2 - X2/ + 2/2 = 6. 20 x + X2/ + 2/ = 5, ■ a;2 + X2/ + 2/2 = 7. Hint. Eliminate x2 or 1/2 as if _ . . o « the equations were linear equa- 21. "^ » ^ i'j tions in x2 and y^. X2/ = 2 x - 2/ + 9. 22 «2_x2/ + 2/2 = 37, (x + 2/)(8-x) = 10, • x2 - j/2 = 40. ■ (X + 2/) (5 - 2/) = 20. 24 (»2 + 2/2)(x + 2/)=6, 25 (^ + 2/)^ = 3x2 - 2, ' «y(x + 2/) = - 2. * • (X - 2/)2 = 32/2 - 11. SIMULTANEOUS QUADRATIC EQUATIONS 117 26. X + y/x^y = a, y + ^x?/2 = 6. 28. X + y = 58, Vx + Vy = 10. 30. X + 2/ = 3. 32. 4x*-9y2 = o, ^ —O ■ ..O O / 1 -A 34. 38. 40. 42. 4x2 + 2/2 = 8(x + y). 3x2/-2(x + y) = 28, 2x2/-3(x + 2/) = 2. x2 + ?/2 + X + y = 18, x2 - 2/2 + X - y = 6. 3x2- 2 2/2 = 6(x-2/), X2/ = 0. x2-X2/ + 2/2 = 13(x-2/), xy = 12. Vx(l-y) +V2/(l-ic) = a, 2/ ^^ Vl-x2Vl-2/2 + X2/ = |l, • x-y = ll. X _y _ 16 46. y X " 16' 3x2 + 6y2 = i20. X V^ + 2/ Vy _ 1 48. xy/x — yy/y 2 x3 - 8 = 8 - y3. V2/ — Va — X = Vy -X, 50. Vy — X + Va — x _ 5 52. Va — X 1 , 1_3 i + i = l x2 y« 4 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. xVx + y = 3, y Vx + y = 1. ■y/x +y/y = a, X + y = 6. Vx — Vy = 2, (x + y) Vxy = 610. Vx-6 + Vy + 2 = 5, X + y = 16. xy + xy-i = x2 + y2, xy-xy-i = 2(x2 + y2). x-2 + 2 y-2 = 12, x-2 - x-iy-i + y-2 = 4. 5c2 + y2_5(a; + y)^8, x2 + y2-3(x + y) = 28. 2x2-3xy + 6y- 6 = 0, (x-2)(y-l) = 0. V6-3x + x2 + V6-3y + y2 = 6, X + y = 3. X y ' y = .3. a y _ 25 y '^x~12* x2 - y2 = 28. V 5x /x + a; + y \ 6a icy - (x + y) = 1. ^ + ?^ = 2, a2 62 bx + ay _m bx — ay n x8 _ y3 _ 16 y X 2 ' - -1=-. y x ~ 2' y ^ 3 V2 X 2 ' Hint. Let - = u, - = v. a ' y 118 QUADRATICS AND BEYOND X -1 a—1 2, 54. ) ^i 55. 3. y-\ b- x3 - 1 a3 yS _l 53 _ 1 lift ^ + ^ 57 ^^^ ^ 2 x + l 221 1 1_5 1_?-1 5g x'^y~6' 59 "^ ^~ ' X + 22/ + I2 x + xy = 2. PROBLEMS 1. Two numbers are in the ratio 5 : 3. Their product is 735. What are the numbers ? 2. Divide the number 100 into two parts such that the sum of their squares is 5882. 3. The sum of the squares of two numbers increased by the first is 205 ; if increased by the second the result is 200. What are the numbere ? 4. The diagonal of a rectangle is 85 feet long. If each side were longer by 2 feet, the area would be increased by 230 square feet. Find the length of the sides. 5. The diagonal of a rectangle is 89 feet long. If each side of the rec- tangle were 3 feet shorter the diagonal would be 85 feet long. How long are the sides ? 6. The sum of two numbers is 30. If one decreases the first by 3 and the second by 2 the sum of the reciprocals of the diminished numbers is ^. What are the numbers ? 7. The sum of the squares of two numbers is 370. If the first were increased by 1 and the second by 3, the sum of the squares would be 500. What are the numbers? 8. A number of persons stop at an inn, and the bill for the entire party is $24. If there had been 3 more in the party, the bill would have been $33. How many were in the party and how much did each pay ? SIMULTANEOUS QUADRATIC EQUATIONS 119 9. A fruit seller gets $2 for his stock of oranges. If his stock had con- tained 20 more and he had charged f of a cent more for each, he would have received $3 for his stock. How many oranges had he and how much did he get apiece for them ? 10. A man has a rectangular plot of ground whose area is 1250 square feet. Its length is twice its breadth. He wishes to divide the plot into a rectangular flower bed, surrounded by a path of uniform breadth, so that the bed and the path may have equal areas. Find the width of the path. 11. In going 7500 yards one of the front wheels of a carriage makes 1000 more revolutions than one of the rear wheels. If the wheels were each a yard greater in circumference, the front wheel would make 625 more revolutions than the rear wheel. What is the circumference of the wheels ? 12. A man has |539 to spend for sheep. He wishes to keep 14 of the flock that he buys, but to sell the remainder at a gain of $2 per head. This he does and gains $28. How many sheep did he buy and at what price each? 13. A man buys two kinds of cloth, brown and black. The brown costs 25 cents a yard less than the black, and he gets 2 yards less of it. He spends $28 for the black cloth and $25 for the brown. How much was each a yard and how many yards of each did he get ? 14. A man left an estate of $54,000 to be divided among 8 persons, namely, his sons and his nephews. His children together receive twice as much as his nephews, and each one of his children receives $8400 more than each one of his nephews. How many sons and how many nephews were there ? 15. A and B buy cloth. B gives $9 more for 60 yards than A does for 45 yards ; also B gets one yard more for $9 than A ddes. How much does each pay? 16. A sum of money and its interest amount to $22,781 at the end of a year. If the sum had been greater by $200 and the interest :^ of 1 per cent higher, the amount at the end of the year would have been $23,045. What was the sum of money and what was the interest ? 17. If one divides a number with two digits by the product of its digits, the result is 3. Invert the order of the digits and the resulting number is in the ratio 7 : 4 to the original number. What is the number ? 18. What number of two digits is 4 less than the sum of the squares of its digits and 5 greater than twice their product ? 19. Increase the numerator of a fraction by 6 and diminish the denomi- nator by 2, and the new fraction is twice as great as the original fraction. Increase the numerator by 3 and decrease the denominator by the same, and the fraction goes into its reciprocal. What is the fraction ? 120 QUADRATICS AND BEYOND 121. Equivalence of pairs of equations. In the theorems of this section the capital letters represent polynomials in x and y, and the small letters represent numbers not equal to zero. I Theorem I. The pairs of equations A are equivalent. If (ari, 2/1) be a pair of values that satisfy (1), then when x and y in 5^ are replaced by Xi and yi the equation B"^ = i^ is a numer- ical identity. These values (xy, y^ must then satisfy one of the equations -S = ± ^, for if they did not, but only satisfied the equa- tion say B = c when c =^ ±b, then the hypothesis that B^ = }p- is satisfied by (cci, ?/i) would be contradicted. Conversely^ any pair of values that satisfy B = -^h evidently satisfy B"- = b\ This theorem is used, for instance, in exercise 2, p. 114, and justifies the assumption that are equivalent pairs of equations. Theorem II. The pairs of equations are equivalent. li A = a and B = h are satisfied by a pair of numbers (xj, y^), we multiply the identities and obtain AB = ab. Conversely f if A = a, AB = ab are identically satisfied by a pair (xi, yi), since a ^ we can divide the second identity by the first and obtain B = b. Thus if (cci, yi) satisfy one pair of equations they satisfy the other pair. This theorem is assumed in exercise 3, p. 115, to show that a;« + w« = 12 ^ , a;2 - XM + 7/2 = 6 1 . , ^ , ^ y and . " ~r ^ ^ are equivalent. SIMULTANEOUS QUADRATIC EQUATIONS 121 Theorem III. The pairs of equations ^:^}a) ^f!^!:^K^) B = 0]^ ' cA-\-dB = are equivalent where a, b, c, and d are numbers such that ad — be ^ 0. If (xi, yi) satisfy (1), evidently they also satisfy (2). Thus all solutions of (1) are among those of (2). Conversely J if (xi, ^/i) satisfy (2), then __bB__dB a c Thus (ad — hc)B = 0. Thus since (ad — bo) =^ 0, . ^ = 0. Similarly, ^ = 0. This theorem has been assumed in exercises 1, 2, 3, 6, p. 114. In 1, for example, it is necessary to show that are equivalent. In this case a=c=l, b = — d = 2. Thus ad — be = — 4: ^t 0. 122. Incompatible equations. When a pair of simultaneous equations can be proven equivalent to a pair of equations which contradict each other or are absurd, they are incompatible and have no finite solution. (1) (2) Example 1. xy = 1 Subtract, xy = -1 0= 2 Example 2. x'^ + y^ = 4, 4x2 + 42/2 = 49. Multiply (1) by 4, Subtract, 4x2 + 42/2 = 16 4x2 + 42/2 = 49 = 33 122 QUADRATICS AND BEYOND 123. Graphical representation of simultaneous quadratic equa- tions. Every equation that we have considered may be rep- resented graphically by plotting in accordance with the method already given (p. 93). The solution of simultaneous equations is represented by the points of intersection of the cor- responding graphs. Thus the equations x'^,f = 25, 2xy = 9 have the solutions a;=±2± V34 y==F2± V34 or X = 4.9, .9, - .9, - 4.9, y .9, 4.9, - 4.9, - .9. These equations have as their graph the preceding jfigure. The equations ic2 + 2£cH-4?/ + l = 0, x-f22/ + 4 = 0, which have the solutions a; = ± Vt = ± 2.6, y = - 2 =F V^ = - 3.3 or - Y'^ - ^ - ^^ ^N, X Z '^ ^J\ - / ^\r - r %> 7 L^ !-—- 5: r> ~fcM — ^^ ^ , Ny 1 /T "\ V I ^ ^ I^ -\ ^ ° -4 t* -^^^- J t _ ^^ 7 "^ have as their graph the figure shown above. As in the case of linear equations, incompatible equations afford graphs which do not intersect. Thus the graph of the equations in example 2, p. 121, is found to be two concentric circles, as is shown in the adjacent figure. SIMULTANEOUS QUADRATIC EQUATIONS 123 Simultaneous equations which have imaginary solutions also lead to non-inter- secting graphs (p. 101). Thus the equations a;2 + 2/' = 4, have the adjacent figure as their graph. . .^.- . ^, V : ■ /"=VlXI 1 1 1 >L f J hs' ^^^/| 1 1 U" EXERCISES 1. Inteipret the graphical meaning of equivalent pairs of equations. 2. Plot and solve x^ + ?/2 = 2, X + 2/ = 2. What general statement concerning the graphical meaning of a single solution of quadratic and linear equations does this example suggest ? 3. Plot and solve the following : (a) 25. X2 + 2/2 4x2 + 9^2 = 144. (b) (c) x2 + 2/2 = 25, x2 + y2 = 25, 4x2 -8x + 92/2 = 140. 5x2 + 2/2 = 25. What general statement concerning the graphical interpretation of four, three, or two real solutions of equations do these examples suggest ? 4. State the algebraical condition under which two quadratic equations have four, three, two, or one real solutions (see p. 113). 5. Plot and solve the following : x2 + 2/ = 0, „ . a;2 + 2/2 = 9. 32/ = 0. (a) (c) (e) X2 4x-22/ = 3, ^.2 I 2/2 y^' x2 - y2 = 0. xy 0. ^ ' X2 + y2 16. 4x2 + 92/2 = 36, x2 = — 4 ?/. (f) xy = 1, 2x-Sy = lS. 124. Graphical meaning of homogeneous equations. Consider for example the homogeneous equation 3x2 -10x2/ -82/2 = 0. (1) If we let z = - , we get X or or 3 _ 10 z - 8 z2 = 0, 8^2 + 102-3 = 0, (4z-l)(2z + 3) =0. 124 QUADRATICS AND BEYOND The roots are 2 = 1 and 2 = - |. Thus y=i and y=-l, a; 4 X 2 4y-x = and 3x + 22/ = 0. or These equations represent two straight lines through the origin which taken together form the graph of equation (1). This example may obviously be generalized : Any homo- geneous equation of the form ox^ + hxy + cy"^ = with positive discriminant represents two straight lines through the origin. Such an equation is equivalent to two linear equations. In an example like 5, p. 115, we obtain in place of the given pair of equations a pair of equivalent equations one of which is homogeneous and the other of which is factorable. We can learn the graphical meaning of tl;is method of solution by studying a particular case. Consider for example the equations : jc2 + 2x2/ + 7?/2 = 24, (1) 2x^-xy-y^ = S. (2) By eliminating the constant terms we obtain the product of the two equations x + y = and x — 2 y = 0. Thus the problem of solving (1) and (2) is replaced by that of solving the two following pairs of equations : x2 + 2X2/ + 72/2 = 24, (^) (2) . (2) x + y = 0, or x2 + 2x2/ + 73/2 = 24, X - 2 2/ = 0. The graphical meaning of this method of solving the equations (1) and (2) is seen in the fact that the problem of finding the points of inter- section of the graph of equation (1) with that of (2) is changed to that of finding the intersection of the graph of (1) with a pair of straight lines. This appears in the figure where the curves and lines are numbered as above. The closed curve represents (1). CHAPTEE XI MATHEMATICAL INDUCTION 125. General statement. Many theorems are capable of direct and simple proof in special cases, while for the general case a direct proof is difficult and complicated. If we ask whether ic" — 1 is divisible by x — 1, it is easy to make the actual division for any particular value of n, as n = 2 or n = 3. But if x^ — 1 is shown divisible by a; — 1, we are no wiser than before concerning the divisibility of x^ — 1. Suppose, however, we can prove that the divisibility for ti = m -f 1 follows from that for n = m, whatever value m may have. Then since we have established the fact by direct division for n = 3, we may be assured of the divisibility for ?i = 4, then for n = 5, and so on. Now x'^+^ — l=x(x"' — l)-\-(x — l) is identically true. If ic — 1 is a factor of cc"* — 1 for a given value of m, it is a factor of the right-hand member and consequently a factor of the left-hand member (§ 69), which was to be proved. Thus the divisibility of x"" — 1 hjx — 1 is established for any integral value of n greater than the one for which the division has actually been carried out. To complete the proof of a theorem by mathematical induction we must make two distinct steps. ^-^^^st Establish the theorem for some particular case or cases, preferaUy for n =1 and n = 2. Second. Show that the theorem for n = m -{- 1 follows from its assumed validity for n = m. Example. Prove that the sum of the cubes of the integers from 1 to 71 is SK^(^ + 1)]S'- To prove that 1« -f- 2» 4- 3* + • • • -F 7i» = J^[7i(n + 1)]^. 125 126 QUADRATICS AND BEYOND First. This theorem is true for n = 1. For 1» = 1 = J ^ [1 (2)] p = 12 = 1 The theorem is also true for n = 2. For l» + 2» = 9= J^[2(2 + l)]P = G-6)2=32 = 9. Second. Assume the theorem for n = m,* 1« + 2« + . . . + w« = J J[m(m + 1)]P. Add (m + 1)^ to both sides of the equation, 1« + 2« + . . • + m« + (m + 1)« = ^[m(m + 1)] p + (m + 1)* = [(^m)2-|-m + l](m + l)2 = ( 4 ) (^ + ly = [i(mH-l)(m + 2)]S which is the form desired, i.e. m + 1 replaces m in the formula. EXERCISES Prove by mathematical induction that 1. 1+3+6 + .. .+(2n-l)=n2. 2. 2 + 22 + 28 + . . . + 2" = 2(2» - 1). 3. 3 + 6 + 9 + ... +3n = f n(n + l). 4. 12 + 22 + 32 + ... + n2 = ^n(n + l)(2n + l). 5. 13 + 28 + 33 + . . . + n3 = (1 + 2 + 3 + . . . + n)2 6. 42 + 72 + 102 + . . . + (3 n + 1)2 = ^ n(6 n2 + 16n + 11). 7. X" — y" is divisible by x — y for any integral values of n. 8. x2» — 2/2" is divisible by x + y for any integral values of n. 9. 1.2 + 2-3 + 3. 4 + 4.6 + . .. + n.(n + l) = ^n(n + l) (n + 2). 10. 1 • 1 + 2 • 32 + 3 . 62 + ... + n(2 n - 1)2 = ^n(n + l)(6n2 - 2n - 1). 11. 1.2.3+2.3.4+3.4.6 + ...+n(n+l)(n+2) = in(?i+l)(n+2)(n+3). 12. (1« + 28 + 38 + • . . + n8) + 8(16 + 26 + 35 + . . . + n6) = 4(1 + 2 + 3+ ... +n)8. * This statement does not imply that we assume the validity of the formxila for any values for which it has not yet been established, but only for values of m not greater than 2. MATHEMATICAL INDUCTION 127 14. Ill 1 _ ^ 1.2'2.3'3.4' 'n-(w + l) n + 1 15. o 16. 2.6 + 3.6 + 4.7 + ... + (« + l)(. + 4) = "<» + ''g><'' + «V 17. 2.4 + 4.6 + 6.8 + -.. + 2n(2n + 2) = ^(2n + 2)(2n + 4) o 18. A pyramid of shot stands on a triangular base having m shot on a side. How many shot are in the pile ? CHAPTEE XII BINOMIAL THEOREM 126. Statement of the binomial theorem. When in previous problems any power of a binomial has been required we have obtained the result by direct multiplication. We can, however, deduce a general law known as the binomial theorem, which gives the form of development of (a + hy, where n is any positive integer and a and h are any algebraical or arithmetical expressions. This law is as follows : (a 4- ^»)« = a« + ^ a«- ^b + ^ ' ^f ~ ^^ a^-^'b^ + "- + b\ From this expression we deduce the following EULE FOR THE DEVELOPMENT OF (a + bf. The first term is a". n 1 To obtain any term from the preceding term, decrease the exponent of a in the preceding term by 1 and increase the exponent of b by 1 for the new exponents. Multiply the coefficient of the preceding term by the exponent of a, and divide it by the exponent of b increased by 1 for the new coefficient. Remark. In practice it is usually more convenient first to write down all the terms with their proper exponents, and then form the successive coefficients. EXERCISES Verify by multiplication the rule given for the following : 1. (a + 6)». 2. (x - 2/)8. 3. (2a + 36)*. 4. (v^+Vi^)'. 5 (2 a -6)4. 6. {x-y/y)\ 7. (3a -2 6)8. 8. (a-ia; + 6-iy)4. 128 BINOMIAL THEOREM 129 127. Proof of the binomial theorem. We have already stated that / -j \ {a + ^)« = a" + 7 a--^b + ""^ ~ ^ a^-'b"" + ••. (1) and have seen that it is justified for every particular case that we have tested. By complete induction we now prove this theorem when ti is a positive integer. First. Let n = 2. That is, (a -{- by= a^ -{- 2 ab + b\ This expression evidently obeys the law as stated in (1). Second. Assume the theorem for n = m. That is, (a + &)- = a- 4- ^ a^-^h + '^^'^ -^) ^m-2^2 + .... (2) Multiply both members hj a -\-b, (a 4- Z>)"'+i = a'«+i ^ja'^h^j a'^-^b'^ z Z This expression is identical with (2) except that (m + 1) replaces m. Hence the theorem is established so far as the first three terms are concerned. 128. General term. Though we have stated the binomial theorem for a>§eneral value of ri, we have only established the exact form of the first three terms. Let {a + by = a'* + c^a^'-^b + Cga"" ^'^ ^ . , We note that the sum of the exponents of a and 5 is w in any term of the development of {a -\- by. Also the exponent of b in the (r -\- l)st term is r. We have already seen that n n(n — l) and that the first three terms are J. ± • z 130 QUADKATICS AND BEYOND respectively. This indicates that the coefficient of the next term will be — ^^ — - — -^r-^ and in general that the coefficient of the (r + l)st term has the form _ n(n-l)---(n-r + l) l-2...r ' W which is in fact the form that our rule (§ 126) would afford for any particular value of r. This affords the following Rule. The (r + l)st term of {a + hf is n{n-l)-"{n-r + l) ^„_ , ^, 1.2'- r The form of the coefficient may be easily remembered since the denominator consists of the product of the integers from 1 to r, while the numerator contains an equal number of factors consist, ing of descending integers beginning with n. For any particular values of n and r we could easily verify the rule by direct multiplication. For the rigorous proof see p. 178. EXERCISES Develop by the binomial theorem : /_a_ _ ^^Y Solution 6-5-4-3 (_±y(_ VxV 6.5.4.3-2 / a V/ VaV 1.2.3.4VV^/V a2/ l-2.3.4.6VVi/\ aV 6. 5-4. 3.2.1 / V^y 1.2.3.4.5.6V a2/ x« x2 X a3 a6 a* "^ a^a" 2. (f-fa)6. 3. (Va + v'ft)'. BINOMIAL THEOREM 131 10. (1 + V^y - (1 - V^)'. 11. ( Vx + v^)*+ ( Vx - Vy)*. (2 X 3 v\^^ 1 ) • Sy 2x7 Solution : n = 10, r + 1 = S. The (r + l)st term of (a + 6)" is (§ 128) n(n-l)-'-(n-r + 1) 1-2. -r In this case we get 10-9-8.76-5.4 (^xy/Syy 1.2.3.4-6-6.7 '\3y)\2x) 33y3 27x7 24X* _ 120-81 y^_ 1215y4 16 'x*~ 2x* ' (1\13 a + -) . 14. Find the 6th term of (— - ^V^. \2y xj 15. Find the 8th term of ( — - ^ ) . \y X / 16. Find the 6th term of l2aVb ) . V 2aVb/ 17. Find the 7th term of (^ - ^) • 18. Find correct to three decimal places (.9)^ Solution; (.9)8 = (1 - .1)8 8^7^6^ ^ 1.2.3.4^ / V ' -r = 1 - 8 • 0.1 + 28 • 0.01 - 66 . 0.001 + 70 • 0.0001 = 1 + 0.28 + 0.0070 - 0.8 - 0.056 = 1.2870 - 0.856 = .431. In this exercise any terms beyond those taken would not affect the first three places in the result. 132 QUADRATICS AND BEYOND Compute the following correct to three places : 19. (1.1)10. 20. (2.9)8. 21. (.98)11. 22. (1.01)«^ 23. (1^)8. 24. (1^)10. 25. (98)8. 26. (203)5. 27. In what term of (a + 6)2o does a term involving ai* occur? 28. For what kind of exponent may a and h enter the same term with equal exponents ? 29. For what kind of exponent is the number of terms in the binomial development even ? 30. Find the first three and the last three terms in the development of (Z, . 1 \24 CHAPTEB XIII ARITHMETICAL PROGRESSION 129. Definitions. A series of numbers such that each numbei minus the preceding one always gives the same positive or negative number is called an arithmetical series or arithmetical progression (denoted by A.P.). The constant difference between any term and the preceding term of an A.P. is called the common difference. The series 4, 7, 10, 13, • • • is an A.P. with the common difference 3. The series 8, 62, 5, 32, • • • is an A.P. with the common difference — §. The series 4, 6, 7, 9, 10, • • • is not an A.P. EXERCISES Determine whether the following series are in A.P. If so, find the common difference. 1. 6, m If, .... 2. 27, 22^, 18, .... 3. 6, 4^ 3, H, .... 4. 6, -2, -8, .... 5. VI, V2, 3 VI, •••. 6. 8, 5|, 3|, If, .... - 1 2 4 V2-I V2 1 '• V2 V2 V2 2 2 2(V2-1) * 9.3, -^, -3f, -6f, .... 10.^ ^^ + 2 -^^ 6 6(V3-4) 130. The nth. term. The terms of an A.P. in which a is the first term and d the common difference are as follows: a, a -i- d, a -\- 2d, a -{- 3dj '". (1) The multiple of d is seen to be 1 in the second term, 2 in the third term, and in fact always one less than the number of the term. If we call I the nth. term, we have I = a -{■ (^n — 1) d. 133 ^ .., a 134 QUADRATICS AND BEYOND We may also write the series in which I is the nth term as* follows : a, a-{- dy a + 2d, • • • , I — 2d, I — d, I. 131. The sum of the series. We may obtain a formula for computing the sum of the first n terms of an A.P. by the following Theorem. The sum s of the first n terms of the series a, a -^dy • • •, I — d, I is By definition, s = a-\-(a + d) + (a + 2d)-\ -]-{l - 2d)-\-(l - d)+ I. (1) Inverting the order of the terms of the right-hand member, s =. I + {l - d) + {I - 2 d) -^ ' ■ - + {a + 2 d) + (a -\- d)+ a. (2) Adding (1) and (2) term by term, 2s^{l-\-a)+{l + a)-\-{l + a) + .'-^{l-\-a) + (l-\-a) + {l + a) z=n{a-\- 1). Thus s = ^(a-\-l). 132. Arithmetical means. The terms of an A.P. between a given term and a subsequent term are called arithmetical means between those terms. By the arithmetical mean of two numbers is meant the number which is the second term of an arithmetical series of which they are the first and third terms. Thus the arithmetical mean of two numbers a and h is — r— > since the numbers a, — ^r— > h are in A.P. with the common difference — — • The two formulas « = a + (»i - 1) d, (I) « = -(« + 1) (II) contain the elements a, I, s, n, d. Evidently when any three are known the remaining two may be found by solving the two equa- tions (I) and (II). AKITHMETICAL PROGRESSION 135 EXERCISES 1. Find the 16th term and the sum of the series 4, 2, 0, — 2, • • .. Solution: n = 16, a = 4, d = 2 - 4 = - 2. Z = a + (n - l)d = 4 -t- 15(- 2) = -26,^ « = |(«4-0 = f(4-26)=-176. 2. Z = 42, a = - 3, d = 3. Find n and s. 3. a = - 4, n = 8, s = 64. Find d and I. 4. d = - i, n = 6, « = 21. Find s and a. 5. d = — i, n = 10, s = 65. Find a and i. 6. s = 161, Z = 4, a = — 3. Find d and n. 7. Z = 22, s = 243, n = 13. Find a and d. 8. s = - 15, Z = - 2, d = 2. Find n and a. 9. d = 41, a = - 16, s = 140. Find n and i. 10. Insert 8 arithmetical means between 4 and 28. 11. Find expressions for n and s in terms of a, I, and d. 12. Find expressions for I and a in terms of s, n, and d. 13. Find expressions for a and s in terms of d, /, and n. 14. Find expressions for d and n in terms of s, a, and Z. 15. Find the 13th term and the sum of the series V2-1 V2 1 2 ' 2 '2(V2-l)'"" 16. Find the 10th term and the sum of the series V3 3V3 + 2 V3 2 T"'~~6~"'"2" + 3'"- 17. Insert 4 arithmetical means between — - and V2 2 10\/6 18. Insert 6 arithmetical means between -x - and 4 19. Insert 3 arithmetical means between and 2 20. Find the 21st term and the sum of the series , V^ V2 V2 21. Find the 10th term and the sum of the series — = 136 QUADRATICS AND BEYOND 22. Eind expressions for d and a in terms of s, Z, and n. 23. Find expressions for d and I in terms of a, n, and s. 24. Find the 8th term and the sum of the series x, 4ic, 7x, • • •. 25. Find the 9th term and the sum of the series 8, 9J, 10|, • • •. 26. Find the 12th term and the sum of the series 8, 7y\, 6|, • • • . 27. Find the 8th term and the sum of tlie series — 8, — 4, 0, • • •• 28. Find the 12th term and the sum of the series 27, 22 1, 18, • • •. 29. Find the 20th term and the sum of the series 1, — 2i, — 6, • • •. 30. Find the 11th term and the sum of the series 5, — 3, — 11, • • •. 31. Find the 9th term and the sum of tlie series x — y, x, x -]- y, • • -. * 32. Insert n — 2 arithmetical means between a and I. Write the first tliree. Kemark. Often an exercise may be solved more simply if instead of assum- ing the series x, x-\- y, z + 2y, ■■ -we assume x — y, x, x -\- y when three terms are required, or x — 2y, x — y, x, x + y, x + 2y when five terms are required, or x~-3y,x — y,x + y,x-i-3y when four terms are required. 33. The sum of the first three terms of an A. P. is 15. The sum of their squares is 83. Find the sum of the series to ten terms. 34. Find expressions for n and a in terms of s, Z, and d. For what real values of s, Z, and d does a series with real terms not exist ? 35. In an A. P. where a is the first term and s is the sum of the first n terms, find the expression for the sum of the first m terms. 36. Find expressions for n and I in terms of a, s, and d. For what real values of a, s, and d does a series with real terms not exist ? 37. If each term of the series (1), § 130, is multiplied by m, is the new series in A. P., and if so, what are the elements of the new series ? 38. If each term of the series (1) in § 130 is increased by 6, is the new series in A. P., and if so, what are the elements of the new series ? 39. The difference between the third and sixth terms of an A. P. is 12. The sum of the first 10 terms is 45. Find the elements of the series. 40. Find the 10th term of an A. P. whose first and sixteenth terms are 3 and 48. Find also the sum of those eight terms of the series the last of which is 60. 41. Two A.P.'s have the same common difference, and their first terms are 2 and 4 respectively. The sum of the first seven terms of one is to the sum of the first seven terms of the other as 4 is to 6. Find the elements of both series. 42. The three digits of a number are in A. P. The number itself divided by the sum of the digits is 48. The number formed by the same digits in reverse order is 396 less than the original number. What is the number? CHAPTEE XIV GEOMETRICAL PROGRESSION 133. Definitions. A series of nTimbers such that the quotient of any term of the series by the preceding term is always the same is called a geometrical progression (denoted by G.P.). The constant quotient of any term by the preceding term of a G.P. is called the ratio. The G.P. series 4, 8, 16, • ■ • has the ratio 2. The G.P. series 8, 4 Vi, 4, • • • has V9 the ratio J-^i . ^' EXERCISES Determine which of the following series are in G.P. and find the ratio. 1. 4, 2, 1, .... 3. 8, -2, .5, .... 5. Vl» i» Vl, •••• 2. 4, 8, 16, .... 4. 8, -4, -2, .... 6. 6,-21, 73i, ... 8. ^ 2 ^ V2' '' V2' 10. V3 [3" V3 8 ' V32' 4 ' 7.-^, A 2, .... V2 \ 6 V5 Vl5 11. —J^ — -^, 5-2V6,3V3-V2,.... 12. V2 - 1, 1, V2' + 1, .... V3 - V2 134. The nth term. The terms of a G.P. in which a is the first term and r the ratio are as follows : a, ar, ar^^ at^, .... The power of r in the second term is 1, in the^ third term is 2, and in fact is always one less than the number of the term. If we call I the nth term, we have the following expression for the T^th term : 137 ^ 138 QUADRATICS AND BEYOND 135. The sum of the series. We obtain a formula for finding the sum of the first 7i terms of a G.P. by the following Theorem. The sum s of the first n terms of the geometrical progression a, ar, ar^^ ... is a — rl .{■> ' = T^r- " By definition, s = a -\- ar + ar^ H- • • • 4- ar"~^ = a(l -h r + r"^ -\ h r^~^) K^) a — rar"^ ^ a — rl by (III), p. 15 by (I), p. 137 136. Geometrical means. The n — 2 terms between the first and the ?ith term of a G.P. are called the geometrical means between those terms. If one geometrical mean is inserted between two numbers, it is called the geometrical mean of those numbers. Thus the geometrical mean between a and h is ^ ah. The two fundamental formulas « = ar»*-i, (I) _ a{\-r^) _ a-rl ^ - 1-r - "rr7 ^^^^ contain the five elements a, I, r, n, s, any two of which may be found if the remaining three are given. EXERCISES 1. Find the 7th term and the sum of the G.P. 1, 4, 16, . . .. Solution : a = 1, n = 7, r = 4. Substituting in (I), I = ar*'-^ = 1 • 4^ = 4096, a — rl 1-r o V *-^ *• • /TTv 1-4-4096 1638.3 ^,^, Substitutmg m (II), s = = = 6461. GEOMETRICAL PROGRESSION 139 2. Insert 2 geometrical means between 4 and 32, 3. Insert 4 geometrical means between 32 and 1. 4. Insert 3 geometrical means between 3 and |f . 5. Insert 4 geometrical means between a^ and l^. 6. Insert 4 geometrical means between 1 and 9V3. 7. Insert 3 geometrical means between y- a-iid 73|. 8. What is the geometrical mean between 3 and 27 ? 9. Insert 3 geometrical means between VS and "v/24. 10. Insert 4 geometrical means between a and a^ Va6^. 11. Insert 3 geometrical means between — | and — 2^. 12. What is the geometrical mean between — 2 and — f ? 13. Find the 7th term and the sum of the series 1, 3, 9, • • •. 14. Find the 6th term and the sum of the series 2, 4, 8, • • • . 15. What is the geometrical mean between -y/a^h and VoP ? 16. Find the 7th term and the sum of the series 8, 2, .5, • • •. 17. Find the 8th term and the sum of the series ^^j, i^, j, • • • . 18. Find the 7th term and the sum of the series Vi, 2, 2V2, • • •. 19. Find the 7th term and the sum of the series \/2, V^, Vi, • • • . 20. Find the 10th term and the sum of the series ■^\-^^ y^^, ^^j, • • •. 21. Find the 5th term and the sum of the series V2 — 1, 1, 1 + V2, • • •. 22. The first and sixth terms of a G.P. are 1 and 243. Find the interme- diate terms. 23. Find the 5th term and the sum of the series ^ :, 6-2V6, 9V3-11V2, ••.. V3+ Vii 24. Insert 3 geometrical means between — and - . V3 9 25. What is the geometrical mean between ~ and ^^ ^? X -\- y X -y 1 - 4 26. Find the 6th term and the sum of the series , — •\/2, — =, • • .. V2 V2 27. Find the 6th term and the sum of the series -* /- , 1, — --, • • •. \3 V2 28. Find the 5th term and the sum of the series v^, — 1, , • • •. 29. Find the 6th term and the sum of the series , -» / — , , • • •. 8 \32 4 140 QUADRATICS AND BEYOND 30. The geometrical mean of two numbers is 4 and their sum is 10. Find the numbers. 31. The fourth term of a G.P. is 192, the seventh term is 12,288. Find the first term and the ratio. 32. If the same number be added to or subtracted from each tern^^^^f^ G.P., is the resulting series geometrical? 33. The product of the first and last of four numbers in G.P. is 64. Their quotient is also 64. Find the numbers. 34. The product of four numbers in G.P. is 81. The sum of the second and third terms is i. Find the numbers. 35. If every term of a G.P; be multiplied by the same number m, is the resulting series a G.P.? If so, w^hat are the elements? 36. The sum of three numbers in G.P. is 42. The difference between the squares of the first and the second is 60. What are the numbers ? 37. The difference between two numbers is 48. The arithmetical mean exceeds the geometrical mean by 18. Find the numbers. 38. Four numbers are in G.P. The difference between the first and the second is 4, the difference between the third and the fourth is 36. Find the numbers. 39. A ball falling from a height of 60 feet rebounds after each fall one third of the last descent. What distance has it passed over when it strikes the ground for the eighth time ? 40. The difference between the first and the last of three terms in G.P. is four times the difference between the first and second terms. The sum of the numbers is 208. Find the numbers. 41. An invalid on a certain day was able to take a single step of 18 inches. If he was each day to walk twice as far as on the preceding day, how long before he can take a five-mile walk ? 42. The difference between the first and the last of four numbers in G.P. is thirteen times the difference between the second and third terms. The product of the second and third terms is 3. Find the numbers. 137. I nfinite series. When the number of terms of a G.P. is unlimited it is called an infinite geometrical series. In the series a, ar, ar^^ • • •, when r > 1, evidently each term is larger than the preceding term. The series is then called increas- ing. When r < 1, each term is smaller than the preceding term and the series is called decreasing. T,T . a (1 — r") a ar^ Now m any case s = — \ = 1 — r 1 — r 1 — r GEOMETRICAL PROGRESSION 141 When r > 1, evidently r" becomes very large for large values of n. For this case, then, the sum of the first n terms becomes very large for large values of n. In fact we can take enough terms so that s will exceed any number we may choose. If, however, r < 1, as 71 increases in value r" becomes smaller and smaller. In fact we can choose n large enough so that r" is as small as we wish, or as we say, approaches as a limit. But since r" may be made as small as we wish, ar^ also approaches as a limit, and conse- quently approaches as a limit. Thus when r < 1 the value of the sum of the first n terms approaches as n 1 — r becomes very great. This we express in other words by asserting that the sum of the infinite series a -\- ar -\- ar^ + ■ "j when r < 1, is s^ = 1-r . EXERCISES Find the sum of the following infinite series. 1. 6 + 3 + 1 + .... Solution : a =z 6, r = |. a 1-r 6 = 5 = 12. i-i h 3. 64 + 8 + 1+.. 5. ^ + i^ + :rV + - 7. 2 + .5 + .125 + 2. l + i + i + .... 4. h + \ + l + -"' 6. ! + f + -V + --- ■v/2 8. V2 + l + -y- + .... 9. (V2 + l) + l + (v^-l) + .... 10. How large a value of n must one take so that the sum of the first n terms of the following series differs from the sum to infinity by not more than .001? (a) 8 + 4 + 2 + .... Solution : a = 8, r = |. a ar'* ar** 8 = = Sao — 1-r 1-r 1 8^-8= 1-r 142 QUADRATICS AND BEYOND We must find for what value of n the expression is less than .001. say 8-2 2» 16 2«* '"^ - - 16 1 ^ By trial we see that if w = 14 the value of — is , which is less than .001 2". 1^2^ (b) 27 + 3 + i + . . .. (c) 4 + I + ^ij + • • •. (d) 1 + ^1^ + 3,1^ + .... (e) 64 + 16 + 4 + .... (f) 100 + 20 + 4 + . . •. (g) 60 + 20 + 6f% . . .. 11. What is the value of the following recurring decimal fractions? (a) .212121.... Solution : This decimal may be written in the form Here (b) .333.... (e) .343343 21 21 21 100 ' (100)2 ' (100)3"'" • 21 1- 100 ' 100 5 _ « _ .21 _ .21 _ 7 * 1-r l-.Ol .99 33* (c) .717171.... (d) .801801.. (f) 1.43131.... ^ (g) 2.61414.. ADVANCED ALGEBRA • CHAPTEE XV PERMUTATIONS AND COMBINATIONS 138. Introduction. Before dealing directly with the subject of the chapter we must answer the question, In how many distinct ways may two successive acts be performed if the first may be performed in p ways and the second may be performed in q ways ? Suppose for example that I can leave a certain house by any one of four doors, and can enter another house by any one of five doors, in how many ways can I pass from one house to the other ? If I leave the first house by a certain door, I have the choice of all five doors by which to enter the second house. Since, how- ever, I might have left the first by any one of its four doors, there are 4 • 5 = 20 ways in which I may pass from one house to the other. This leads to the v,^ Theorem. If a certain act may he 'performed in p ways, and if after this act is performed a second act may he performed in q ways, then the total numher of ways in which the two acts may he performed is p • q. With each of the p possible ways of performing the first act correspond q ways of performing the second act. Thus with all the p possible ways of performing the first act must correspond q times as many ways of performing the second act. That is, the two acts may be performed in ^ • 5' ways. It is of course assumed in this theorem that the performance of the second act is entirely independent of the way in which the first act is performed. 143 144 ADVANCED ALGEBRA EXERCISES 1. I have four coats and five hats. How many different combinations coat and hat can I wear ? Solution : The first act consists in putting on one of my coats, whicli may be done in four ways-, the second act consists in putting on one of my hats, which may be done in five ways. Thus I have 4 • 5 = 20 different combinations of coat and hat. 2. In how many ways may the two children of a family be assigned to five rooms if they each occupy a separate room ? 3. A gentleman has four coats, six vests, and eight pairs of trousers. In how many different ways can he dress ? 4. I can sail across a lake in any one of four sailboats and row back in any one of fifteen rowboats. In how many ways can I make the trip ? 5. Two men wish to stop at a town where there are six hotels but do not wish quarters at the same hotel. In how many ways may they select hotels ? 6. A man is to sail for England on a steamship line that runs ten boats on the route, and return on a line that runs only six. In how many different ways can he make the trip ? 7. In walking from A to B one may follow any one of three roads; in going on from B to C one has a choice of five roads. In how many different ways can one walk from A to C ? 139. Permutations. Each different arrangement either of all or of a part of a number of things is called a permutation. Thus the digits 1, 2 have two possible permutations, taken both at a time, namely, 12 and 21. The digits 1, 2, 3 have six different permutations when two are taken at a time, namely, 12, 13, 21, 23, 31, 32. For if we take 1 for the first place, we have a choice of 2 and 3 for the second place, and we get 12 and 13. If 2 is in the first place, we get 21 and 23. Similarly, we get 31 and 32. In this process it is noted that we can till the first of the two places in any one of three ways ; the second place can be filled in each case in only two ways. Thus by the Theorem, § 138, we should expect 3-2 = 6 permuta- tions of three things taken two at a time. We observe that this product 3 • 2 has as its first factor 3, which is the total number of things considered. The number of factors is equal to the number of digits taken at a time, i.e. two. This leads to the general PEKMUTATIONS AND COMBINATIONS 145 Theorem. Tlie number of permutations of n ohjects taken r at a time is , i\ , , -t\ /t\ n(n — 1) ■ ■ ' (n — r -\- J^)' v-j This is symbolized by P„^ ^. This formula is easily remembered if one observes that the first factor is 7i, the total number of objects considered, and that the number of factors is r, the number of objects taken at a time. Thus ■Py3 = 7-6-5. We prove this theorem by complete induction. Flrstj let r = 1. There are evidently only n different arrange- ment of n objects, taking one object at a time, namely (assuming our objects to be the first 7i integers), 1, 2, 3, ..-, 71. Let us take two objects at a time, i.e. let r= 2. Since there are n objects, we have n ways of filling the first of the two places. When that is tilled there are n — 1 objects left, and any one may be used to fill the second place. Thus, by the Theorem, § 138, there are for r = 2 . . . n {n — V) different permutations. Secondy assume the form (I) for r = m^ Pn,m = n(n-l)..-(n-m + l). (1) We can fill the first m places in P„ ^ different ways since there are that number of permutations of n things taken m at 'a time. This constitutes the first act (§ 138). The second act consists in filling the m -f 1st place, which may be done in n — m ways by using any of the remaining n — in objects. Thus the number of permutations of n things taken 7n + 1 at a time is Pn,r,^ + i = Pn,nr(n-m)=:=n(n-l){n-2)---(n-m + l)(n-m\ which is the form that (1) assumes on replacing m by m + 1. Corollary. The number of permutations of n things taken all at a time is ^ , -,\ ^ -t » * /o\ F^^^=n(n-'l)-'-2-l = n!* (^) Taking n = r in (I), we get (2). » « / is the symbol for 1 • 2 .3 • 4 •••(»- 1) n, and is read factorial n. 146 ADVANCED ALGEBRA EXERCISES 1. How many permutations may be formed from 8 letters taken four at a time? Solution : n = 8, r=»4, n-r + l=6, Pg, 4 = 8 • 7 . 6 . 5 = 1680. 2. In how many different orders may 6 boys stand in a row ? 3. How many different numbers less than 1000 can be formed from the digits 1, 2, 3, 4, 5 without repetition ? 4. How many arrangements of the letters of the alphabet can be made taking three at a time ? 5. How many numbers between 100 and 10,000 can be formed from the digits 1, 2, 3, 4, 5, 6 without repetition ? 6. How many different permutations can be made of the letters in the word compute taking four at a time ? 7. In a certain class there are 4 boys and 5 girls. In how many orders may they sit provided all the boys sit on one bench and all the girls on another ? Hint. Use Corollary § 139, and then Theorem, § 138. 8. I have 6 books with red binding and 3 with brown. In how many ways may I arrange them on a shelf so that all the books of one color are together ? 140. Combinations. Any group of things that is independent of the order of the constituents of the group is called a combination. The committee of men Jones, Smith, and Jackson is the same as the committee Jackson, Jones, and Smith. The sound made by striking simultaneously the keys EGrC of a piano is the same as the sound made by striking CGE. In general a question involv- ing the number of groups of objects that may be formed where the character of any group is unaltered by any change of order among its constituent parts is a question in combinations. Suppose for example that we ask how many committees of three men can be selected from six men. If the men are called A, B, C, D, E, F, there are, by § 139, 6 -5 • 4 = 120 difPerent arrangements or permutations of the six men in groups of three. But the permu- tations A, B, C ; A, C, fe ; B, A, C, etc. (3 ! = 6 in all for the men A, B, and C), are all distinct, while evidently the six committees consisting of A, B, and C are identical. This is true for every distinct set of three men that we could select; that is, for the PERMUTATIONS AND COMBINATIONS 147 six different permutations of any three men there is only one distinct committee. Hence the number of committees is one sixth the total number of permutations, or -^' This leads to the general Theorem. The number of comhinations of n things taken r at a time is . -,^ , . ^x n(n — I)--- (n — T-\- 1) This is symbolized by C„^ ^. The number of permutations of n things taken r at a time is p^^^ = n(n - 1) • ' • (r, — r + 1). In every group of r things which form a single combination there are (Cor., p. 145) r ! permutations. Thus there are r ! times as many permutations as combinations. That is, r -?Ji^- n{n-l)-'-{n-r-\-l) "''•""/•!" r\ ' ^^ This formula is easily remembered if one observes that there is the same number of factors in the numerator as in the denomi- nator. Thus 10- 9- 8 ^10.3- -^.2-3 Corollary. C ^= C n, n — r Multiplying numerator and denominator of (I) by (n — r)!, _ n(n-l)-'-(n-r + l)(n-r)-'-2-l "•'•" rl(n-ry. _ n{n-l)---{r-\-l) (n — r)\ — ^ (^ ~ 1) • • • [^ ~ (^ ~ ^) + ^] (n — r)\ n.n — r' This corollary saves computation in some cases. For instance, if we wish to compute Ci9, ir, it is more convenient to write Ci9, 17= Ci9, 2 = ^ , = 171 than the expression for C'lo, 17. 148 ADVANCED ALGEBRA EXERCISES 1. How many committees of 5 men can be selected from a body of 10 men three of whom can serve as chaii'man but can serve in no other capacity ? Solution : There are 7 men who may fill 4 places on the committee. ^' 1.2.3-4 There are 3 men to select from for the remaining place of chairman, and the selection may be made in 3 ways. Thus the committee can be made up in 3 • 35 = 105 ways. 2. How many distinct crews of 8 men may be selected from a squad of 14 men ? 3. How many distinct triangles can be drawn having their vertices in 10 given points no three of which are in a straight line ? 4. How many distinct sounds may be produced on 9 keys of a piano by striking 4 at a time ? 5. In how many ways can a crew of 8 men and a hockey team of 5 men be made up from 20 men ? 6. In how many ways may the product a-b • c • d- e -f be broken up into factors each of which contains two letters? 7. If 8 points lie in a plane but no three in a straight line, how many straight lines can be drawn joining them in pairs? 8. How many straight lines can be drawn through n points taken in pairs no three of which are in the same straight line ? 9. Seven boys are walking and approach a fork in the road. They agree that 4 shall turn to the right and the remainder turn to the left. In how many ways could they break up ? Solution : The number of groups of 4 boys that can be formed from the CV 4 = = o5. 4! For each group of 4 boys there remains only a single group of 3 boys. Thus the total number of ways in which tliQ party can divide up is precisely 35. 10. If there are 12 points in space but no four in the same plane, how many distinct planes can be determined by the points? Hint. Three points determine a plane. 11. Eight gentlemen meet at a party and each wishes to shake hands Tfith all the rest. How many hand shakes are exchanged ? I PERMUTATIONS AND COMBINATIONS 149 12. In how many ways can a baseball team of 9 men be selected from 14 men only two of whom can pitch but can play in no other position ? 13. How many baseball teams can be selected from 15 men only four of whom can pitch or catch, provided these four can play in either of the two positions but cannot play elsewhere ? 14. Two dormitories, one having 3 doors, the other having 6 doors, stand facing each other. A path runs from each door of one to every door of the other. How many paths are there ? 15. Show that the number of ways in which p -{- q things may be divided into groups of p and q things respectively is ^ ^^• p\q\ 16. Out of 8 consonants and 3 vowels how many words can be formed each containing 3 consonants and 2 vowels ? 17. A boat's crew consists of 8 men, three of whom can row only on one side and two only on the other. In how many ways can the crew be arranged ? 18. A pack of cards contains 62 distinct cards. In how many different ways can it be divided into 4 hands of 13 cards each ? 19. Five points lie in a plane, but no three in any other plane. How many tetrahedrons can be formed with these points taken with two points not in the plane ? 141. Circular permutations. By circular permutations we mean the various arrangements of a group of things around a circle. Theorem. The number of orders in which n things may he arranged in a circle is (n—l) !. Suppose A is at the point at which we begin to arrange the digits 1, 2, 3, • • •, n. Suppose we start our arrangement of digits at A with a given digit a. We have then virtually n—l places to fill by the remaining n—l digits. Thus we get {n—l)\ (p. 145) permutations of the n digits keeping a fixed. But suppose we start our arrangement, that is, fill the place at A with any other digit, as h, and the remaining places in any order what- ever. If we now go around the circle till we come to the digit a, the succession of digits from that point around the circle to a again must be one of the {n —1)\ orders 150 ADVANCED ALGEBRA which we obtained when we took a as the initial figure. Thus the only distinct orders in which the n digits can be arranged on a circle are the (n —1)\ permutations we obtained by filling the first place with a. EXERCISES 1. In how many orders can 6 men sit around a circular table ? Solution : n = Q, n - 1 = 5, (n - 1)! = 5 ! = 120. 2. In how many ways can 8 men sit around a circular table ? 3. In how many ways may the letters of live be arranged on a circle ? 4. In how many ways may the letters of permutation be arranged on a circle ? 5. In how many ways can 4 men and 4 ladies sit around a table so that a lady is always between two men ? 6. In how many ways may 4 men and their wives be seated around a table so that no man sits next his wife but the men and the women sit alternately ? 7. In how many ways can six men and their wives be seated around a table so that each man sits between his wife and another lady ? 8. In how many ways can 10 red flowers and 5 white ones be planted around a circular plot so that two and only two red ones are adjacent ? 142. Theorem. The number of permutations of n things of . n! which p are alike, taken all together y is -^• If all the things were different, we should have n ! permutations. But since p of the n things are alike, any rearrangement of those p like things will not change the permutation. Eor any fixed arrangement of the n things there are p ! different arrangements of the p like things. Thus — : of the n\ permutations are iden- tical, and there are only — '- distinct permutations of the n things ^p of which are alike. ' Corollary. If of n things p are of one kind, q of another n f kind, r of another, etc., then there are — ; — -^— — permutations of the n things taken all at a time. -^ ' ^' ' ^ PERMUTATIONS AND COMBINATIONS 151 EXERCISES 1. How many distinct arrangements of the letters of the word Cincinnati are possible ? Solution : There are in all 10 letters, of which 3 are i, 2 are c, and 3 are n. Thus the number of arrangements is 10! _ X.2.^-^-5-^-7-8-9.10 3!3!2!" ;.^.3X-^-^;-;Z = 2.5-7.8.9.10 = 50,400. 2. How many distinct arrangements of the letters of the word parallel can be formed ? 3. How many signals can be made by hanging 15 flags on a staff if 2 flags are white, 3 black, 5 blue, and the rest red ? 4. How many signals can be made by the flags in exercise 3 if a white one is at each extreme ? 5. How many signals can be made by the flags in exercise 3 if a red flag is always at the top ? 6. Would 3 dots, 2 dashes, and 1 pause be enough telegraphic symbols for the letters of the English alphabet, the numerals, and six punctuation marks ? CHAPTEE XVI COMPLEX NUMBERS 143. The imaginary unit. When we approached the solution of quadratic equations (p. 52) we saw that the equation x^ = 2 was not solvable if we were at liberty to use only rational num- bers, but that we must introduce an entirely new kind of number, defined as a sequence of rational numbers, if we wished to solve this equation. The excuse for introducing such numbers was not that we needed them as a means for more accurate measurement, — the rational numbers are entirely adequate for all mechanical purposes^ — but that they are a mathematical necessity if we propose to solve equations of the type given. A similar situation demands the introduction of still other numbers. If we seek the solution of £C2=-1, (1) we observe that there is no rational number whose square is — 1. Neither can we define V— 1 as a sequence of rational numbers which approach it as a limit. We may write the symbol V— 1, but its meaning must be somewhat remote from that of V2, for in the latter case we have a process by which we can extract the square root and get a number whose square is as nearly equal to 2 as we desire. This is not possible in the case of V^l. In fact this symbol differs from 1 or any real number not merely in degree but in kind. One cannot say V— 1 is greater or less than a real number, any more than one can compare the magnitude of a quart and an inch. V — 1 is symbolized by I and is called the imaginary unit. The term "imaginary" is perhaps too firmly established in mathe- matical literature to warrant its discontinuance. It should be kept in mind, however, that it is really no more and no less 152 COMPLEX NUMBERS 153 imaginary than the negative numbers or the irrational numbers are. So far as we have yet gone it is merely that which satis- fies equation (1). When, however, we have defined the various operations on it and ascribed to it the various characteristic properties of numbers we shall be justified in calling it a number. Just as we built up from the unit 1 a system of real numbers, so we build up from V— 1 = i a system of imaginary numbers. The fact that we cannot measure V— 1 on a rule should cause no more confusion than our inability exactly to measure -y/2 on a rule. Just as we were able to deal with irrational numbers as readily as with integers when we had defined what we meant by the four operations on them, so will the imaginaries become indeed numbers with which we can work when we have defined the corresponding operations on them. 144. Addition and subtraction of imaginary numbers. We write = 0i, i -\- i = 2 ij i -{- i-\ -{- i = ni. (I) Also just as we pass from a rational to an irrational multiple of unity by sequences, so we pass from a rational to an irrational multiple of the imaginary unit. Thus we write a V— 1, or ai, where a represents any real number. Consistently with § 76 we write ^ V-a2 = ^ Va2.(_l) = ± V^ . V^ = ±a V^ = ± ai. (II) We speak of a positive or a negative imaginary according as the radical sign is preceded by a positive or a negative sign. We also define addition and subtraction of imaginaries as follows : ai ± bi = (a ±b) i, (III) where a and b are any real numbers. 154 ADVANCED ALGEBRA Assumption. The commutative and associative laws of multi- plication and addition of real numbers, § 10, we assume to hold for imaginary numbers. 145. Multiplication and division of imaginaries. We have already virtually defined the multiplication of imaginaries by real numbers by formula (I). Consistently with § 76 we define V^ . V^ = ii = i^ = - 1. Thus V— a • V— b = Va • V^ i i = -\/ab • (— 1) = — -Vab. The law of signs in multiplication may be expressed verbally as follows : The product of imar/inaries with like signs before the radical is a negative real number. The product of imaginaries with unlike signs is a positive real number. For instance, - yP^ - V^^ = - 2 • 3 • i2 = 6. We also note that i2 = — 1, i3 = — i, i* = 1, i^ =r i, . . . , And, in general, Hn + k _ ik^ /c = 0, 1, 2, 3. We define division of imaginaries as follows : / I r Vo^ • i la ■Vb-i ^b In o perating with imaginary numbers, a number of the form V— a should always be written in the form Va i before per- forming the operation. This avoids temptation to the following error: V- a • V- b = V(- a) ■(-b) = -slab. EXERCISES Simplify the following : 1. V^8 ■ yT^. Solution : V^^ • V^^ = VS • i . V2 • i = V2 • 8 . {2 = 4 . (_ 1) = _ 4. 2.1. 1 2^ /^ ■^— % Solution : — = - = = = — i. i^ i8 (i4)2 \ COMPLEX NUMBEKS 165 3. i". 6. V-36. 9. V- X2«. 12. V2 V- 8. 15. 1 18. V-6 4. 124. 5. ii3. 7. V-64. 8. 2i'Si. 10. V-Sx^a^. 11. V-x2. 13. V-2V-6. 14. V-3% 16.1. 17.. f^. V-2 19. V-i2. 20. V-i^. 146. Complex numbers. The solution of the quadratic equation with negative discriminant (p. 71) affords us an expression which consists of a real number connected with an imaginary number by a H- or — sign. Such an expression is called a complex number. It consists of two parts which are of different kinds, the real part and the imaginary part. Thus 6 + 4 z means G I's + 4 ^'s. Obviously, to any pair of real numbers (x, y) corresponds a complex number x + ty, and conversely. 147. Graphical representation of complex numbers. We have represented all real numbers on a single straight line. When we wished to represent two numbers simultaneously, we made use of the plane, and assumed a one-to-one correspondence between the points on the plane and the pairs of numbers (cc, y). The general complex number x -f- iy depends on the values of the independent real numbers x and y, and may then properly be represented by a point on a plane. We repre- sent real numbers on the X axis, imaginary numbers on the Y axis, and the complex number x + iy by the point (x, y) on the plane. Thus the complex numbers 6 + ^ 3, — 4 -f 1 4, 7 — i 5, — 2 — 1 4 are represented by points on the plane as indicated in the figure. 148. Equality of complex numbers. We define the two com- plex numbers a -f ib and c -{- id to be equal when and only when a = c and h — d. y> k - 4+ i4 6f i?, X -2- i4 7- 15 rn 156 ADVANCED ALGEBRA Symbolically a + ii ^ c + id when and only when a = c, b = d. The definition seems reasonable, since 1 and i are different in kind, and we should not expect any real multiple of one to cancel any real multiple of the other. Similarly, if we took not abstract expressions as 1 and i for units but concrete objects as trees and streets, we should say that a trees + b streets = c trees + d streets when and only when a = c and b = d. Principle. When two numerical expressions involving imagi- naries are equal to each other, we may equate real parts and imaginary parts separately. The graphical interpretation of the definition of equality is that equal complex numbers are always represented by the same point on the plane. From the definition given we see that a -\-ib = when and only when a = b = 0. Assumption. We assume that complex numbers obey the com- mutative and associative laws and the distributive law given in § 10. We also assume the sa.me rules for parentheses as given in § 15. This assumption enables us to define the fundamental opera- tions on complex numbers. 149. Addition and subtraction. By applying the assumptions just made we obtain the following symbolical expression for the operations of addition and subtraction of any two complex num- bers a -{-ib and c -{- id: a -{• ib ± (c -\- id) = a ± c -{- i(b ± d). Rule. To add (subtract) complex numbers, add (subtract) the real and imaginary parts separately. 150. Graphical representation of addition. We now proceed to give the graphical interpretation of the operations of addition and subtraction. COMPLEX NUMBERS 167 Theorem. The sum of two numbers A = a-\- ib and B = c -\- id is represented hy the fourth vertex of the parallelogram formed on OA and OB as sides. Let OASB'hQ ?i parallelogram. Draw ES _L OE, AH A. ES, BD (= d) _L OE. A AHS = A ODE since their sides are parallel, and OB — AS. Thus Thus DB = : HS = d, 0D = AH==c. ES = EH + HS Y 1 d A 7 S H b D —a, > JF E X b + d, OE = OF + FE = a -\- c, and S has coordinates (a -\- c, b -{- d) and represents the sum of A and B, by § 149. EXERCISES 1. The difference A — B of two numbers A = a + ib and B = c + id is represented by the extremity D of the line OD drawn from the origin par- allel to the diagonal BA of the parallelogram formed on OB and OA as 2. Represent graphically the following expressions. (a) 1 + i (b) - 4 - 2 i ■ (c) 6 - i. (d) - 8 + 4 i. (e) 2 + 4 I (f) (1 + i) + (2 + i)' (g) {2-i)-{6-Si). (h) (l_i)-(l_2i). (i) (2 + 4i)-(l-3i). (j) 4(l + i)-2(2-3i). (k) (6-2i) + (2 + 3i). (1) (5 + 3i) + (-l-6i). 151. Multiplication of complex numbers. The assumption of § 148 enables us to multiply complex numbers by the following Rule. To multiply the complex number a -\- ib by c ■\- id, pro- ceed as if they were real binomials, keeping in mind the laws for multiplying imaginaries. Thus a -\-ib c -f- id ac + icb -f- iad -\- (i)* bd = ac — bd + i(cb -f- ad). 158 ADVANCED ALGEBRA 152. Conjugate complex numbers. Complex numbers that differ only in the sign of their imaginary parts are called conjugate com- plex numbers, or conjugate imaginaries. Theorem. The sum and the product of conjugate complex numbers are real numhers. Thus a -\- lb -\- a — ib = 2 a, (a + ib) (a - ib) =a^ + b\ 153. Division of complex numbers. The quotient of two com- plex numbers may now be expressed as a single complex number. EuLE. To express the quotient — in the form x + ^y, rationalize the denominator^ using as a rationalizing factor the conjugate of the denominator. ^_^ ™,, a -{- ib a -\- ib c — id Thus — = ~ ^^—^ c -\- id c -\- id c — id _ ac -\- bd — i {ad — be) ~ c' + d^ ac -{- bd .ad — be ^ > + d' ~ "" VT^' ^^ We have now defined the fundamental operations on complex numbers and shall make frequent use of them. If the question remains in one's mind, "After all, what are they? " the answer is this : " They are quantities for which we have defined the funda- mental operations of numbers and, since they have the properties of numbers, must be called numbers, just as a flower that has all the characteristic properties of a known species is thereby deter- mined to belong to that species." Furthermore, our operations have been so defined that if the imaginary parts of the complex numbers vanish and the numbers become real, the expression defining any operation on complex numbers reduces to one defin- ing the same operation on the real part of the number. Thus in (1) above, if b = d = 0^ the expression reduces to a _a a c COMPLEX NUMBERS 159 EXERCISES Carry out tlie indicated operations. 1. (2 + V"r2)(4+V^^). Solution : 2 + V^^ = 2 + ■v/2(-l) = 2 + z V2 4 + V:^ =:4+V5(-l) = 4 + t V6 8 - VlO + i4V2 + i2\/5 2. 5 - V2 - i Vs. Solution : 5^ ^ 5(V2 + zV3) ^ 5V2 + Z5V3 _ /g + i Vs \/2-iV3 (V2-iV3)(V2 + iV3)~ 2 + 3 3. (l + i)l 4. (l + i)3. Hint. Develop by the binomial theorem. 5. {a + ib)K 6. (V^ + v^ir^)^ V7. (x + %)2 8. (x + i2/)2 + (X - %)2. 9. vT+l.Vnri. 10. (V3 + iv^)(V2 + iV3). 11. (VTT~i + Vr^y. 12. (aV6 + icVd)(aV6-icVd). 13. (Va + i V6) ( Va - i Vb). 14. (2 V7 + i3 Vs) (3 V7 - ilOV2), 15. i±i^. 16. l±i. 17. ^ l-iV3 1-i V2+V-1 18. ^ . 19. 11^. 20. (^l±i^)'. l + V-3 (l + i)3 \ 2 / 2„ g + i Vl - a2 a — i Vl — a^ 24. ;^-^- 25. -^1— _. 26. ^ + ^p^ . 27. ^^ + ^^'^ 28 (l±i^y. 29. ^ + ^^ _j_ c + id V3-iV2 \2/ a-ib c -id 30. ?? 31. — ?i=. 32. -V+ 1 4 + 7V^6 i + 3v^^. (1 + 0' a-*y 33 VI + « + ^ Vl- <^ _ Vl - g + i Vl + g Vi -j- g — i VT — g Vl — a — i Vl + a 160 ADVANCED ALGEBRA 34. Find three roots of the equation x^ — 1 = and represent the roots as points on the plane. 35. Find four roots of the equation x* — 1 = and represent the roots as points on the plane. 36. Find six roots of x^ — 1 = and represent the roots as points on the plane. Show graphically that the sum of the six roots is zero. 37. Find three roots of x^ — 8 = and represent the roots as points on the plane. Show graphically that the sum of the three roots is zero. 154. Polar representation. The graphical representation of complex numbers given in § 147 gives a simple graphical inter- pretation of the operations of addition and subtraction, but the graphical meaning of the operations of multiplication and divi- sion may be given more clearly in another manner. We have seen that we may represent x -f iy by the point P (x, y) on the plane, Represent the angle between OP and the X axis by/^. This angle is called the argument of the complex number x -f iy. Eepresent the line OP by p. This is called ^ , . , the modulus of ic -f iy. Then from the figure a; = p COS B, (1) — ^ y = psmO, (2) x' + f = p\ (3) Hence the complex number x + iy may be written in the form X -\- iy = p (cos -}- i sin 0)j (4) ■when the relations between x, y and p, are given by (1), (2), and (3). A number expressed in this way is in polar form, and may be designated by (p, 6). We observe that a complex number lies on a circle whose center is the origin and whose radius is the modulus of the number. The argument is the angle between the axis of real numbers and the line representing the modulus. 155. Multiplication in polar form. If we have two numbers p (cos 6 -\-i sin^) and /o'(cos 6' -\- i sin ^'), we may multiply them and obtain COMPLEX NUMBERS 161 p (cos 6 + isin 0)p'(Gos 0' + i sin 6') = pp' [(cos cos 0' — sin 6 sin d') + i (sin $ cos d' + cos sin ^')] By the addition theorem ,r //i , /if\ , • • /h , /if\n /-i\ in Trigonometry = PP C^^^ (^ + ^') + ^ sm (d + d')] (1) = R (cos © + * sin ©) . (2) In this product pp' is the new modulus and 0-^0' the new argument. We may now make the following statement: The product of the two numbers p (cos 6 -\- i sin 6) and p'(cos 6' + * sin 0') has as its modulus pp' and as its argument + 0'. Thus the product of two numbers is represented on a circle whose radius is the product of the radii of the circles on which the factors are represented. The argument of the product is the sum of the arguments of the factors. 156. Powers of numbers in polar form. When the two factors of the preceding section (p, 0) and (p', 6') are equal, that is, when p = p' and 6 = 6', the expression (1) assumes the form [p (cos e + i sin 0)^ = p^ (cos 2 ^ + t sin 2 0). (1) This suggests as a form for the nth. power of a complex number [p (cos d -\-i sin $)']" = p« (cos nO -\- i sin nO). (2) The student should establish this expression by the method of complete induction. The theorem expressed by (2) is known as DeMoivre's theorem. Stated verbally it is as follows : The modulus of the nth. power of a number is the nth power of its modulus. The argument of the nth power of a number is n times its argument. EXERCISES Plot, find the arguments and moduli of the following numbers and of eir products. 1. 1 + iVS, V3 + i. Solution : Let ^/S -\- i=p (cos ^ + i sin 5), 1 4- i V3 = /(cos d' + i sin 6^. 2. 30°; Y. B Ay ! i 1 ! Then by (1), (2), (3), § 154, p = 2 ; p' = 1 = 2 sin d, hence 6 = '--- -1— J 1 X 1 = 2 cos d", hence 6' = 60°. = 30°, ^' = 60° 162 ADVANCED ALGEBRA Thus if the product has the form R{cos@ + isin©), we have by § 1( R = ppr = 4, © = ^ + ^' = 90°. 2. l+i,2 + i. 3. (l-i)3. 4. 3 + 3t, 2^iVl2. 5. 2i, l-iV3. '4*'-^)- 7.-1 + 1,-2-2. o , . V2 iV2 ^- '^'2 2 • '-2+ 2 ' 2 + 2 10. [2 (cos 15° + i sin 15°)]8. 11. [i (cos 30° + i sin 30°)]4. 12. [| (cos 120° + i sin 120°)]2. 13. [2 (cos 135° + i sin 135°)]*. 14. [f (cos 180° + isin 180°)] 3. 15. [f (cos 315°+ isin 315°)]2. 157. Division in polar form. If we liave, as before, two com- plex numbers in polar form (p, 6) and (p', $'), we may obtain their quotient as follows. p (cos -{- i sin $) p' (cos 0' -^ isin 0') _ pp' (cos -{- i sin 0) (cos 0' — i sin 0') ~ p'2 (cos ^' + 2 sin $') (cos ^' — i sin ^') ^ pp' [cos (6 - ^0 + i sin (^ - ^')] ~ p'2(cos2^' -f sin^^') Rationalizing, § 152 and § 153, Since sin2 + cos2 6=1, = -^ [cos (0 — $') + i sin (^ — 0')'] = i2 (cos^Vf i sin ©). We may now make the following statement: The, quotient of two complex numbers has as its modulus the quotient of the moduli of the factors, and as its argument the difference of the arguments of t/ie factors. 158. Roots of complex numbers. We have seen that the square of a number has as jts modulus the square of the original modulus, while' the argument is twice the original argument. Thjs would suggest th^-t the square root of any number, as (p, 0), ' : ' - 9 would have Vp as its modulus and - as its argument. Since '^ ^ every yeal number has two square roots, we should expect the same fact to hold liere. Consider the two numbers COMPLEX NUMBERS 163 Vp /^cos I + i sin I j and Vp cos (^ -f 180° j + i sin ( | + 180° j j , where Vp is the principal square root of p(§ 72). The square of the first is (p, ^), by § 155. That the square of the second is the same is evident if we keep in mind the fact that cos (0 + 360°) = cos e and sin (^ + 360°) = sin a Thus V/> (cos e + i sin 6) Vp ( cos - + i SlUv- or Vp + 180° -f j sin/ ^ + 180 ■)]■ The graphs of these two numbers are situated at points sym- metrical to each other with respect to the origin. We may obtain as the corresponding expression for the higher roots of complex numbers the following : ■v^/>(cos^ + ?:sin^)= V^ I ( (9 + /c360°\ . . /^ + A;360' cos I + I sin j_ \ 71 / \ n where for a given value of /i, k takes on the values 0, 1, • • •, t^ — 1, and where VP indicates the real positive 7ith root of p. EXERCISES Perform the indicated operations and plot 1. 2-2V3i--l+i. Solution : Let \-\-i — p (cos + i sin ^), 2 -- 2 V3 i = /(cos e'-\- i sin 6"). Then p = \/12 + 1- = V2, / = V22 + (-2V3)^ = 4. By (1) and (2), § 154, sin ^ = cos ^ = — ^ , hence 6 = 46°. V2 '4; '^-^ (a-iaW; 164 Similarly, ADVANCED ALGEBliA 300° sin^ = - 2V3 4 V3 2 cos^ = f = 1, hence d' - 2 _ 2 V3 i 4 (cos 300° + i sin 300°) Thus = i2(cos© + isin©)=— 7^- -— . . ,^^ ■ 1+i ^ ' V2(cos46°4- tsin45°) Hence by § 157, iJ = -^ = 2v^, © = 300° - 45° = 255» V2 / 2. V-2 + 2V3i Let -2 + 2 V3i = p(cos^4'isin^). Then (§ 154) /> = 4, cos^= - f = -^, -2+i2A^ and 120^= V-2 + 2V3i = V4 (cos 120° + i sin 120°) T5 «i^Q /if /120° + A:360°\ By §158, =V4|cos( ^ j l+iVs" . . /120° + A;360°\'] .sm( ^)J + ^ (where A; = or 1) = 2 (cos 60° + i sin 60°) = 1 + i V3, when k-0. = 2 (cos 240° + i sin 240°) = - 1 - i V3, when A; = I . 3. VV2 + i V2. 4. V 1 + i -^ 1 - i. 1 V3 6. 1 iV3 2 2 ^1 + i. 8 -l-i-^-^l + l N 2 2 ■ 4 4* 10. -2\/2-2 V2i-T--2 + 2V3i 7. i + i^:t_j:Jl 2 2 9. 2-iVl2-4-3 + 3i. 11. \^l(cosl5°+isinl5°). Solution : "/TT ^:,o . - - -. ^Q^ '/tF /15° + fe • 360°\ . . . /16°+ fc • 360°\ 1 Vl(cosl5°+ ism 15°) = VI cosi ■ ) + ism I 1 (where fc = 0, 1, or 2). 1 (cos 5° + t sin 5°), when A; = 0, 1 (cos 125° + i sin 125°), when A; = 1, 1 (cos 245° + i sin 245°), when A; = 2. 12. Yi. 13. ^^161. 14. •V'2-f- 2\/3i. 15. v'cos330°+ tsin330<». COMPLEX KUMBERS 165 16. V27(cos75°+ism75°). 17. Vl6 (cos 200° + i sin 200°). 18. Solve the following equations and plot their roots. (a) x5 - 1 = 0. Solution : x^ = 1, or a; = VT. Let 1 = 1 + • i = p (cos ^ + i sin 6). Then p = l, 6 = 0°. X = V 1 (cos 0° 4- i sin 0°) = Vircos/* 0° + A; . 360' (where k takes on the values 0, 1, 2, 3, 4) f cos 0° + i sin 0° = 1, when k = 0, cos 72° + i sin 72°, when A; = 1, cos 144° + i sin 144°, when fe = 2, • cos 216° + t sin 216°, when A; = 8, cos 288° + i sin 288°, when A; = 4. These numbers we observe lie on a circle of unit radius at the vertices of a regular pentagon. (b) a;4 - 1 = 0. (e) x6 - 1 = 0. J (C) X3-1 = 0. (f ) X8 - 1 = 0. C?x~^ <Jtf^.C<. (d) x5 - 32 = 0. (g) x8 -- 27 = 0. J^l^^ y. O^ CHAPTEE XVII THEORY OF EQUATIONS 159. Equation of the nth degree. Any equation in one variable in whicli the coefficients are rational numbers can be put in the form f (x) =aoX'^ + a^x--' -}--'• + a, = 0, (1) where Gq is positive and a^, • • •, a„ are all integers. The symbol /(x) is read "foix" and is merely an abbreviation for the right- hand member of the equation. Often we wish to replace x in the equation by some constant, as a, — 2, or 0. We may symbolize the result of this substitution by/(a),/(-2),or/(0). Thus f{b) = a(fi» + aift"" ^ + ^ • • + a„ . We symbolize other expressions similarly by <f) (x), Q{x), etc. When we speak of an equation we assume that it is in the form of (1). This equation is also written in the form a;" + M"-i + ---^„ = 0, (2) where bi = —> b^ = —> •• • K — ~' The 5's are integers only when ai, «2j • • -, «„ are multiples of Qq. 160. Remainder theorem. We now prove the following impor- tant fact. Theorem. When f(x) is divided hy x — c, the remainder is f(x) with c substituted in place of the variable. Divide the equation (1) by x — c. Let R be the remainder, which must (§ 26) be of lower degree in x than the divisor ; that is, in this case, since x — c is the divisor, R must be a con- stant and not involve x at all. Let the quotient, which is of degree n ~ 1 in x, he represented by Q (x). 160 THEORY OF EQUATIONS 167 Then i^ = Q(x)+ -5— X — C ^ ^ X — G Clearing of fractions, f(x)= Q(x){x-g)+R. But since this equation is an identity it is ahvays satisfied whatever numerical value x may have (§ 53). Let X = c. Then /(c) = a,c- + %c"-i + ... + «„= Q (c) (c - c) + i2. But since c — c = 0, Q(c) (c — c) = 0, and R = aoc" + ^ic"- 1 H h a„ =/(c)- Corollary. If c is a root of f(x) = 0, then x — c is a factor of the left-hand member. , For if c is a root of the left-hand member, it satisfies that member and reduces it to zero when substituted for x. Thus by the previous theorem we have, since aoC* H- o^ic"-^ H a^ = R = 0, f{x)=Q(x)(x-c). 161. Synthetic division. In order to plot by the method of § 103 the equation y = a^x"" + a^x""- 1 H \- a^, when the a's are replaced by integers, we should be obliged laboriously to substitute for x successive integers and find corre- sponding values of y, which for large values of n involves con- siderable computation. We can make use of the preceding theorem to lighten this labor. The object is to find, with the least possible computation, the remainder when the polynomial f(x) is divided by a factor of form x — c, which by the preceding theorem is the value of f(x) when x is replaced by c, that is, the value of y corresponding to x = c. For illustration, let f(x) = 2 x^ - 3 a;« -h ?c2 - a: - 9 and c = 2. 168 ADVAIJCED ALGEBKA By long division we have a;-2| 2x^-3a;^H- ^' - x- 9 |2a;« + o^' + 3a; + 5 1 x» + a;2 lx3-2a;2 3a;2- X 3a;2-6x 5a;- 9 5a; -10 + 1 We can abbreviate this process by observing the following facts. Since x is here only the carrier of the coefficient, we may omit writing it. Also we need not rewrite the first number of the partial product, as it is only a repetition of the number directly above it in full-faced type. Our process now assumes the form 1- 21 2-3 + 1-1- 9 |2 + l + 3 + 5 :l-4 + 1 -? / / -2 + 3 ,' -6 + 5 10 + 1 Since the minus sign of the 2 changes every sign in the partial product, if we replace — 2 by + 2 we may add the partial prod- uct to the number in the dividend instead of subtracting. This is also desirable since the number which we are substituting for X is 2, not — 2. Thus, bringing all our figures on one line and placing the number substituted for x at the right hand, we have 2-3 + 1-1- 9[2 + 4 + 2 + 6 + 10 2+1+3+5+ 1 THEORY OF EQUATIONS 169 We observe that the figures in the lower line, 2, 1, 3, 5, up to the remainder are the coefficients of the quotient 2 x^ -{- x^ -\- S x -\- 5. EuLE FOR SYNTHETIC DIVISION. Write the coefficients of the polynomial in order, supplying when a coefficient is lacking. Multiply the number to he substituted for x by the first coeffi- cient, and add (algebraically) the product to the next coefficient Multiply this sum by the number to be substituted for x, add to the next coefficient, and proceed until all the coefficients are used. The last sum obtained is the remainder and also the value of the polynomial when the number is substituted for the variable. 162. Proof of the rule for synthetic division. This rule we now prove in general by complete induction. Let the polynomial be a^x"" 4- a^x^-^ + a^x""-^ -\ h «„. Let the number to be substituted for x be a. First. Let n = 2. Carry out the rule on a^x"^ + a^x + a^. We have , , . + aQ(X + {apa + a\) cc <^oi + «o«^ + «i, + {a^a -|- a.i) a + ^2 = a^a^ + a^a, + a^. Second. Assume the validity of the rule for n = m, and prove that its validity f or ti = m + 1 follows. Assume then that the rule carried out on f{x) = a^x^ + a^x^-^ + ••• + «„ affords the remainder a^a"^ + ai^"*-! ^ -\- a^ =f((x)' Now the polynomial of order w + 1 is ^0^"'+^ 4- aix"" H \-a^x-i- a^ + j = x •f(x) + a^^^ Hence the next to the last remainder obtained by applying the rule to this polynomial would be f(cc), since the succession of coefficients is the same for both polynomials up to a^+i- By the rule the final remainder is obtained by multiplying the expres- sion just obtained, in this case f(cc), by a and adding the last coefficient, in this case a^_^,^. This affords the final remainder 170 ADVANCED ALGEBRA EXERCISES 1. Prove by complete induction that the partial remainders up to the final remainder obtained in the process of synthetic division are the coefficients of the quotient of f{x) hj z — a. 2. Perform by synthetic division the following divisions. (a) a;3 - 7x2 _ 6x + 72 by a; - 4. Solution : 1-7-6 + 72 [4 4 _ 12 - 72 1 _ 8 - 18 Quotient = a;2_3x- 18. (b) aj8 - 9x + 10 by X - 2. (c) 4x3 - 7 x - 87 by x - 3. (d) x3 + 8x2 - 4x - 32 i3y x-2. (e) x^ + 4x2 - 7x - 30 by x + 3. (f) x8 - 6x2 + 11 X - 6 by X - 1. (g) x* - 16x3 + 86x2 - 176x + 105 by x2 - 8x + 7. Hint. SinceK2— 8a;+7=(a; — 7)(x — 1), divide by x—7 and the quotient by a;— 1. (h) x6 + 1 by X + 1. (i) x9 - 1 by X - 1. (j) x4 + x3 - X - 1 by x2 - 1. (k) x^ - 2x3 - 4x by x - 3. (1) x6 - 2x8 - 4x - 1 by X + 2. (m) 4x3 - 6x2 - 2x - 1 by x - 3. (n) 2x* + 5x3 - 37x2 +44x + 84 by x2 + 6x - 6. 163. Plotting of equations. We can now form the table of values necessary to plot an equation of the type ^0^" + ^i^'"" ^ H 1- «„-i^ + «„ = y. Example. Plot x3 + 4 x2 - 4 = j/. l + 4 + 0-4[l +1+5+5 1+6+6+1 1 + 4 + 0-41-1 -1-3+3 l+S-S-l 1 + 4 + 0-41-2 -2-4+8 1+2-4+4 l_f.4 + 0-4[-_3 -3-3+9 +1-3+5 1 + 4 + 0-41-4 -4+0+0 X V X y -4 - 1 -1 1 + 1 -2 + 4 ;l xo -3 + 6 -4 -4 1+0+0-4 THEORY OF EQUATIONS 171 In this figure two squares are taken to represent one unit of x. A single square represents a unit of y. By an inspection of the figure it appears that the curve crosses the X axis at about x = .8, x = — 1.2, and x = — 3.7. Thus the equation for y = has approximately these values for roots (§ 110). 164. Extent of the table of values. Since the object of plot- ting a curve is to obtain information regarding the roots of its equation, stretches of the curve beyond all crossings of the X axis are of no interest for the present purpose. Hence it is desirable to know when a table of values has been formed extensive enough to afford a plot which includes all the real roots. If for all values of X greater than a certain number the curve lies wholly above the axis, there are no real roots greater than that value of x. By inspection of the preceding example it appears that if for a given value of x the signs of the partial remainders are all positive, thus affording a positive value of ?/, any greater value of X will afford only positive partial remainders and hence only positive values of y. Thus when ■ all the partial rem.ainders are positive no greater positive value of x need he substituted. Similarly, when the partial remainders alternate in sign begin- ning with the coefficient of the highest power of x, no value of x, greater negatively, need be substituted. In plotting, if the table of values consists of values that are large or are so distributed that the plot would not be well propor- tioned if one space on the paper were taken for each unit, a scale should be so chosen that the plot will be of good proportion, that is, so that all the portions of the curve between the extreme roots shall appear on the paper, and the curvatures shall not be too abrupt to form a graceful curve. This was done, for example, in the figure, § 163. EXERCISES Plot and measure the values of the real roots of the equations when y = 0. 1. x8 - 7 X - 6 = y. 2. x3 - 7 X + 5 = y. 3. 7x8 - 9x - 6 = y. 4. x3 - 31 X + 19 = y. 6. x3 - 12x - 14 = y. 6. 4x8 - 13x + 6 = y. 172 abvakced algi:biia 7. a;8 - 12x - 16 = y. 8. x^ - 45x + 152 = y. 9. x* - 2x3 - X + 2 = y. 10. 8x3 _ igxS + 17x - 6 = y. 11. X* - 17x2 + X + 20 = y. 12. x-i - 4x3 + 9x2 - 8x + 14 = y. 13. 18x3-36x2 + 9x + 8 = 2/. 14. x* + 5x3 + 12x2 + 52x- 40 =y. 15. x*-2x3-7x2+19x-10=y. 16. x4-6x3 + 3x2 + 26x-24 = y. 17. 6x4 - 13x3 + 20x2 - 37x + 24 = y. 165. Roots of an equation. In the case of the linear and quadratic equations we have been able to find an explicit value of the roots in terms of the coefficients. Such processes are prac- tically impossible in the case of most equations of higher degree. In fact the proof that any equation possesses a root lies beyond the scope of this book, and we make the Assumption. Every equation possesses at least one root. This is equivalent to the assumption that there is a number, rational, irrational, or complex, which satisfies any equation. 166. Number of roots. We determine the exact number of roots by the following Theorem. Every equation of degree n has n roots. Given the equation /(x) = ao^?" + aiic""^ -\ +- a„ = 0. Let ^i (see assumption) be a root of this equation. Then (p. 166) ic — ^i is a factor of the left-hand member, and the quotient of f{x) by £c — «! is a polynomial of degree n —1. Suppose that aoic" -f- «ia;«- 1 -f ••• + «« = «o(^ - «^i) (ic"" ^ + ^i^c""^ H h ^„- 1). By our assumption the quotient a;"" ^ -f Jicc'*"^ 4- • • • + ^„_ i = has at least one robt, say az, to which corresponds the factor X — a^. Thus f{x) =aQ{x- a^ {x — a^) (a;"-^ -\- CiX"-^ -\ h c^.g). Proceeding in this way we find successive roots and corre- sponding linear factors until the polynomial is expressed as the product of n linear factors as follows : f{x) =ao(x- ai) (x-a2)'"(x- a„) = 0, where the roots are Uu ag, • • • , a„. Remark. This theorem gives no information regarding how many of the roots may be real or imaginary. This depends on the particular values of the coefl&cients. THEORY OF EQUATIONS 173 Corollary. Any polynomial in x of degree n may he expressed as the product of n linear factors of the form x — a, where a is a real or a complex number. It should be noted that the roots are not necessarily distinct. Several of the roots and hence several of the factors may be identical. If f(x) is divisible by {x — aiY, that is, if a^ = a^, we say that oTi is a double root of the equation. Similarly, \i f(x) is divis- ible by {x — oTi)'', a^ is called a multiple root of order r. When we say an equation has n roots we include each multiple root counted a number of times equal to its order. Theorem. An equation of degree n has no more than n distinct roots. Let/(x) = ^0^" H f" ^n = ^ have the roots a-^^, cc2j'"j *».• Write the equation in the form ao(x — ai) • • ' (x — a^) = 0. If r is a root distinct from a^, •••,«:„, it must satisfy the equation and ao(r-aj)"'(r-a^) = 0. Since this numerical expression vanishes one of its factors must vanish (§5). But r =f= ai, thus r — a^ =^ 0. Similarly, no one of the binomial factors vanishes. Thus (§5) «<, — 0, which contradicts the hypothesis that the equation is of degree n. This theorem may also be stated as follows : Corollary I. If an equation a^x" + a^^x"" "^ + \- a^= of degree n is satisfied by more than n values of x, all its coefficients vanish. The proof of the theorem shows that if the equation has n + 1 roots, ao = 0. We should then have remaining an equation of degree n —1, also satisfied hj n +1 values of x. Thus the coeffi- cient of its highest power in x vanishes. Similarly, each of the coefficients vanishes. 174 ' ADVANCED ALGEBRA Corollary II. If two ^polynomials in one variable are equal to each other for every value of the variable^ the coefficients of like ^powers of the' variable are equal and conversely. Let a^x"" + a^x""-^ H h o^„ = ^o^" + ^i^c""^ H h *„ for every value of x. Transpose, {a^ — h^x''-\--'--\-a^ — h^ = ^. By Corollary I, a^ — h^^^ 0, or a^ = b^, ai — bi = 0, or a^ = bi, (^n-K = 0, or a„ = ^>„. 167. Graphical interpretation. The graphical interpretation of the theorems of the preceding section is that the graph of an equation of degree n cannot cross the X axis more than n times. Since each crossing of the X axis corresponds to a real root, there will be less than n crossings if the equation has imaginary roots. 168. Imaginary roots. We now show that imaginary roots occur in pairs. This we prove in the following Theorem. If a -\- ib is a root of an equation with real coeffi- cients, a — ib is also a root of the equation. li a -{- ib is a root of the equation <Xo^" + a^x^' ^ -\- -••-{- a^ = 0^ then X —(a -{- ib) is a factor (p. 166). We wish to prove that X — {a — ib) is also a factor, or what amounts to the same thing, that their product [x —{a-\- ib)'\ [x —{a — lb)'] = [(x — a) — ib"] [(x — a)-\- ib"] is a factor of f(x). Divide f(x) by (x — a)^ + b^ and we get f(x) = Q(x) [(x - af J^b^-]J^rx + r', (1) where r and r' are real numbers. This remainder rx + r' can be of no higher degree in x than the first, since the divisor (x - af + b'' THEORY OF EQUATIONS 175 is only of the second degree (§ 26). Now this equation (1) being an identity is true whatever value is substituted for x, as, for instance, the root of /(x), a -\- ib. Substituting this value for x, we get i f{a Jrib) = = Q{a + ib) \_{a + ib - ay + b^-]^^ r {a -\- ib) + r', or (p. 33) and (p. 3) = + r«^ -f r' + irb, or (p. 156) m + r' = 0, (2) rb = 0. (3) Since b ^ 0, by (3), p. 3, r = 0. Also from (2), r' = 0. > Hence rx -{- r' = 0. Consequently there is no remainder to the division of f(x) by (x — a)^ -\- U\ and hence if a + ib is a root oif(x), a — ib is also a root. Corollary. Every equation of odd degree with real coeffi- cients has at least one real root. The roots cannot all be imaginary, else the degree of the equa- tion would be even by the preceding theorem. 169. Graphical interpretation of imaginary roots. When we plot the equations y = x^ + Ax''-4. (1), Yk y = x^-i-4:x^-l (2), 176 ADVANCED ALGEBRA 7/ = ^^ + 4^2 (3), y = a;8 + 4 x" +1 (4 0. ^ \ / / \ / / >v / r \ / ^ \ \ \ \ \ \ J x^ , — — X we see that corresponding to the increase of the constant term is a corresponding elevation" of the curve with respect to the X axis. In fact in each case the curve is the same, but the value of y is gradually increased. In (1) and (2) we have three real roots, in (3) the curve touches the X axis, and in (4) we have only one real root. As the elbow of the curve is raised and fails to intersect the X axis a pair of roots cease to be real, and since a cubic equa- tion always has three roots, a pair of roots become imaginary. Thus we have the Pkinciple. Corresponding to every elbow of the curve that does not intersect the X axis there is a pair of imaginary roots of the equation. The converse is not always true. It is not always possible to find as many elbows of the curve which do not meet the X axis as there are pairs of imaginary roots. EXERCISES Plot the following equations and determine from the plot how many roots are real. 1. X* - 1 = y. 2. a;6 - 2 = y. 3. x^ - x - 1 = y. 4. x4 + l = y. 5. x* + x + l = ?/. 6. x* + 2x2 + 2 = y. 7. x8 - 3x2 - X + 1 = y. 8. x8 - 2x2 + 4x - 1 = y. 9. 2x8 + 3x2 + 6x + 6 = y. 10. x^ - 3x2 - 4x - 5 = y. THEORY OF EQUATIONS 177 170. Relation between roots and coefficients. If we write the expression (Corollary, p. 173) and multiply the factors, we obtain by equating coefficients of like powers of x (p. 174) relations between the roots and the coeffi- cients. Take for example n = S. x' + Wx^ + hx + b, = (x- 13,) (x - 13,) (X - p,) = x'- (13, -{-/3, + 13,) x' + ()8i/82 + ^2^3 + ft A) X - )8i A^3 = 0. . Hence j^ = _ (/3^ + ^^ + ^3), *2 = Aft + Aft + A A, ^3 = -AAft- This suggests the Theorem. The coefficient of ^""^ is equal to the sum of the roots with their signs changed. The constant term is equal to the product of the roots with their signs changed. In general the coefficient of a?""*" is equal to the sum of all possible products of r of the roots with their signs changed. We prove this theorem by complete induction. First. We have already established the theorem for equations of degree two on p. 106 and for equations of degree three above. Second. Assume the theorem for n =^m. That is, if £c- + ^'1^'"-' + • • • + *«. = (^ - A) (^ - A) • • • (^ - A). (1) we assume that h^, the coefficient of ic'"-'", is the sum of all possi- ble products of r of the numbers — A? — A? ' • '? ~ A- Multiply both sides of (1) hy x — A« + i. Denote the result by ^m+i + b^^x^ + . . . + 5'^^^ = (x- p,)(x ^p,)...(x- ^^ + 0- (2) The term in 0;"'+^-'" in this equation is obtained by multiplying the terms b.x'^-'- and &^_ia;'»+^-'' in (1) by x and - p^+i respec- tively. That is, in (2) b'r = K + b^-,(-(im^O' (3) Now all possible products of r of the quantities — A> —ft* .. •, — A 4.1 may be formed as follows : (1) Neglect — ft + i, and form all possible products of r of those remaining. The sum of 178 ADVANCED ALGEBRA these is b^. (2) Form all possible products of r - 1 of — )8i — Aj • • •> — Pm^ not including — Pm + u and multiply each product ^y — Pm+\- Add all the products obtained. This process, it is observed, is precisely that indicated by (3). Remark. It is noticed that in the rule the signs of the roots are always changed before forming any term. This does not involve any change when r is an even number, but is included in the rule for the sake of uniformity. Corollary. Every root of an equation is a factor of its con- stant term. 171. The general term in the binomial expansion. On p. 129 we gave an expression for the (r + l)st term of the binomial ex- pansion, the validity of which we now establish. In (1), § 170, let ySi = ^2 = • • • = Pn- Denote this common value by — a. The expression (1) becomes, on writing n in place of m, X- + hx""-' + --- + b,=(x + ay. By the theorem in § 170, b^ is the sum of all possible products of r of the negative roots. Since there are a.= n(n — 1) "• (n — r -i-l) ri such products, and since the roots are now identical, we obtain n(n-l)>-'(n-r-{-l) ^^_^^^ rl as the form of the (r + l)st term of the expansion of (x + a)". 172. Solution by trial. Since by the previous corollary every root of an equation is a factor of its constant term, we may in many cases test by synthetic division whether or not a given equa- tion has integral roots. Thus the integral roots of the equation ic* _ 8a;» + 4:ic=» -f 24a; - 21 = (1) must be factors of 21. We try + 1 by synthetic division, ' 1_8h-4 + 24- 2111 -|.1_7_ 3-1-21 1_7_3-|-21 Thus 1 is a root of (1), and the quotient of the equation by "^-^^^ a;»-7x='-3a;-f 21 = 0. (2) THEORY OF EQUATIONS 179 If this equation has any integral root it must be a factor of 21. We try + 3 by synthetic division, 1_7_ 3 + 21|3 + 3 _ 12 - 45 1-4-15-24 Thus 3 is not a root. We try + 7, 1_7_3 + 21|7 + 7 4- - 21 1+0-3 Thus 7 is a root, and the remaining roots of (1) are the roots of a;2 - 3 = 0, that is, X =± V3. Hence the roots of (1) are + 1, + 7, ± V3. EXERCISES Solve by trial : 1. x3 - 7x2 + 50 = 0. 2, x3 - 9x 4- 28 = 0. 3. x3-36x- 91 = 0. 4. x3 + 9x + 26=:0. 5. x8 - 19x + 30 = 0. 6. x3 - 27x - 54 = 0. 7. x8 + 2x2 - 23x + 6 = 0. 8. x3 - 6x2 + 11 X - 6 = 0. 9. x3 - 2x2 - llx + 12 = 0. 10. x3 - 8x2 ^ 19a; _ 20 = 0. 11. x8 + 9x2 + 27x + 26 = 0. 12. x* - 8x3 + 8x2 ^ 40x - 32 = 0. 13. X* - 13x2 + 48x - 60 = 0. 14. x* - 3x3 - 34x2 + 18x +168 = 0. 15. X* +8x3- 7x2 -50x + 48 = 0. 16. x* - 3x3 - 5x2 + 29x - 30 = 0. 17. x* - 6x3 + 13x2 - 30x + 40 = 0. 18. x4-8x3+21x2-34x + 20 = 0. 19. x* - 12x3 + 43x2 - 42x + 10 = 0. 173. Properties of binomial surds. A binomial surd is a number of the form a ± V^, where a and b are rational numbers, and where b is positive but not a perfect square. Though we have not explicitly defined what we mean by the sum of an irrational number and a rational number, we shall assume that we can operate with the binomial surd just as we would be able to operate if b were a perfect square. 180 ADVANCED ALGEBRA Theorem \. If a binomial surd a + V^ = 0^ then a =0 anc b = 0. J,i a-\- V^ = and either a = ov b = 0, clearly both must equal zero. Suppose, however, that neither a nor b equals zero. Then transposing we have a =^ 'Vb, and a rational number would be equal to an irrational number, which cannot be. Hence the only alternative is that both a and b equal zero. Theorem II. If two binomial surds, asa -\- -sib and c + V5, are equal, then a = c and b = d. Let a + V^ = c -f V^. Transposing c, a — c -\- V^ = V5. (1) Square and we obtain (^ci - cy-\-b -\- 2(a - c)Vb = dy or (^a, - cy + b - d -\- 2(a - c) Vb = 0. Thus, by Theorem I, either b = 0, which is contrary to the defi- nition of a binomial surd, or a ~ e = 0, that is, a = c. In the latter case (1) reduces to Vb == Vd, OTb = d, and we have a = c and b = dy which was to be proved. a 4- Vb and a — V^ are called conjugate binomial surds. Theorem III. If a given binomial surd a + Vb is the root of an equation with rational coefficients, then its conjugate is also a root of the same equation. The proof of this theorem, which should be performed in writ- ing by each student, may be made analogously to the proof of the theorem on p. 174. 174. Formation of equations. If we know all the roots of an equation, we may form the equation in either one of two ways (see p. 167 and p. 177). First method. If a^, oc^,-", oc^ are the given roots, multiply together the factors x — a^,- ■•,x — a^. Second method. From the given roots form the coefficients by the rule on p. 177. THEORY OF EQUATIONS 181 If the equation and all but one of its roots are known, that root can be found by the solution of a linear equation obtained from the coef&cient of the second or the last term. If all but two of its roots are known, the unknown roots may be found by the solution of a pair of simultaneous equations formed from the same coefficients. In the solution of the following exercises use is made of the theorem on p. 174, Theorem III, p. 180, and the various relations between the roots and the coefficients. EXERCISES 1. Form the equations which have the following roots. Check the process by using both methods of § 174. (a) 2, - 3, 1. Solution : First metfiod. {x - 2){x + 3) (x - 1) = JC^ - 7 x + 6. Second method. Let the equation be x3 + 6ix2 + biX + 63 = 0. Then, by § 170, 61 = - (2 - 3 + 1) = 0, - • \ 62 = -6 + 2-3 = -7, 63 = -2-3. -1 = 6. The equation then is x^ — 7x + 6 = 0. (b) 1, 2, 3. (c) 2, 2, 2, 2. (d) 3, 1, 1, 0. (e) 1, 0, 0, 0. (f) ±V2,±i. (g) 2,4, -6. (h) 2, - 3, 1, 0. (i) 2, 3, - 6. . (j) 7, V5, -V5. (k) 1,2, -1, -|. (1) 3, 1 + i, 1 - i (m) _ 4, - 3, 3 ± V5. (n) 1 ± i, - 1 ± i (o) 2, V=^, -V^^. (p) -1,2,3,-4. (q)2^,3|, -H, -2^. (r) ±V6, ±iV7. (s) -5, 2 + V6, 2-V5. (V) 3, ^i±^, i:!^. (w) - 1, l±i^^ 1^. 182 ADVANCED ALGEBRA 2. The equation x* + 2x3 - 7^2 _ 8x + 12 = has two roots — 3 and + 1. Find the remaining roots. Solution : Let the unknown roots be a and h. Then, by § 170, _a-6 + 3-l = 2, - 3 a6 = 12. Solving for a and 5, we obtain a = — 2 or +2, 6 = + 2 or - 2. 3. x8 — 7x + 6 = has the roots 2 and 1. Find the remaining root. 4. X* — 3x + 2 = has the root 1. Find the remaining roots. 5. x^ — 18 X — 35 = has the root 5. Find the remaining roots. 6. Two roots of x* — 35 x^ -f- 90 x — 56 = are 1 and 2. Find the remain- ing roots. 7. The roots of x^ - 6 x2 - 4 x + 24 = are in A. P. Find them, 8. The two equations x3-6x2 + llx-6 = and x^ -14x2 + 63 x - 90 = have a root common. Plot both equations on the same axes, and find all the roots of both equations. 9. Determine the middle term of the equation whose roots are — 2, + 1, 3, — 4 without determining any other term. 10. What is the last term of the equation whose roots are — 4, 4, ± V— 3 ? 11. One root of x* - 4x8 + 5x2 + 2x + 52 = is 3 - 2i. Find the remain- ing roots. 12. One root of x* - 4 x^ + 5 x2 + 8 x - 14= is 2 + i V3. Find the others. 13. Plot the following equations, determine all the integral roots, and find the remaining roots by solving. (a) X* - 6x8 + 24x - 16 = 0. X y -1 - 33 -2 -3 + 155 In this plot two squares on the X axis represent a unit of x, while one square on the Y axis represents ten units of y. The integral factors are X — 2 and x + 2, since ± 2 are roots, that is, are values of x for which the THEORY OF EQUATIONS 183 curve is on the X axis. To find the quotient of our equation we first divide synthetically by 2, and then the quotient by — 2, using the principle given in § 161. 1-6+ + 24- 16 12 + 2 - 8-16 + 16 ^ 1-4- 8+ 8 |-2 - 2 + 12 - 8 1-6+ 4 Thus the quotient of the polynomial and (x — 2) (« + 2) is «2 __ ga; 4. 4 Solving the equation x2-6x + 4 = 0, we obtain the two remaining roots, x = 3 ± V5. These remaining roots might also be found by the method of exercise 2. (b) x3 - 6x - 12 = 0. (c) x3 - 8x2 + 7 = 0. (d) x8 - 7x2 + 50 = 0. (e) x^ - 8x2 + 13x- 6 = 0. (f) x8-6x2+7x-2 = 0. (g) x3 + 3x2 + 4x- 24 = 0. (h) X* - 3x3 + 7 x2 - 21 X = 0. (i) X* - 3x3 - 7 x2 + 27 x - 18 = 0. '(j) x*-9x3 + 21x2-19x + 6 = 0. (k) How many imaginary roots can an equation of the 5th degree have ? (1) x3 — ax2 + 6x + c = has two roots whose sum is zero. What is the third root ? What are the two roots whose sum is zero ? (m) x3 + x2 + 6x + c = has one root the reciprocal of the other. What are the values of the roots ? (n) x3 — 4x2 + ox + 62 = has the sum of two roots equal to zero. What must be the values of a and h ? (o) X* - 3 x3 + 6x + 9 = has the sum of three of its roots equal to zero. What must be the value of 6 ? 175. To multiply the roots by a constant. Suppose we have the equation fix) = a,x- +- a^x--' + . . . + «^ = 0, (1) whose roots are «ri, a^, • • • , «:„. An equation of this type for values of n greater than 2 is usually not solvable by elementary methods. It often happens, however, that by changing its form slightly we may obtain an equation one or more of whose roots we can find. We shall see that if an equation has rational roots we may always find them if we change the form of the equation as indicated on the following page. / 184 ADVANCED ALGEBRA We seek to form from (1) an equation whose roots are equal to the roots of (1) multiplied by a constant factor, as k. Thus the equation we seek must have the roots kai, ka^, ka^. We carry out the proof, which is perfectly general, on the equation of the third order f(x) = a^x^ -f- a-^x^ -f «2^ + ^3 = 0, whose roots are a^, a^, a^. The equation that we seek must have roots kai, ka^, ka^. Since now (§ 53) f{x) = is satisfied by a, where a stands for any one of the. roots, that is, since f(a) = 0, evidently /(t) = is satisfied by ka, that is. Hence we obtain an equation that is satisfied by ka^, ka^^ ka^, if in/(£c) we let a; = t* The required equation is then Akj-'W^l^^-k^''^-^^ or, multiplying by k^, a^z^ + ka-^z'^ -{- k^azZ -{- k^a^ — 0. This affords the general KuLE. To multiply the roots of an equation hy a constant k, multiply the successive coefficients beginning with the coefficient of x""'^ hy k,l^, • • -,1^ respectively. In performing this operation the lacking powers of x should be supplied with zero coefficients. Example. Multiply the roots of 2 x^ — 3 x + 4 = by 2. Multiply the coefficients by the rule above, 2«8 + 2-0x2-4-3x + 8-4 = 0. Simplifying, »» - 6 x + 16 = 0. THEORY OF EQUATIONS 185 When an equation in form (2), p. 166, has fractional coefficients, an equation may be formed whose roots are a properly chosen multiple of the roots of the original equation and whose coeffi- cients are integers. Corollary I. When k is a fraction this method serves to divide the roots of an equation hy a given number. Corollary IL When k = — 1 this method serves to form an equation whose roots are equal to the roots of the original equa- tion hut opposite in sign. This is equivalent to the statement that /(— x) = has roots equal hut opposite in sign to those • EXERCISES 1. Form the equation whose roots are three times the roots of Solution : Supplying the missing term in the equation, we have «* -6x3 + 0x2- X + 1 = 0. Since A; = 3, we liave by the rule x* -3 -6x3 + 9 -0x2- 27 -x + 81 = 0, 'r or x*-18x3 -27x + 81 = 0. ' 2. Find the equation whose roots are twice the roots of x* + 3x3-2x + 4 = 0, 3. Find the equation whose roots are one half the roots of x3-2x2 + 3x-4 = 0. 4. Find the equation whose roots are two thirds the roots of x3-4x-6 =0. 5. By what may the roots of the following equations be multiplied so thai in the resulting equation the coefficient of the highest power of x is unity and the remaining coefficients are integers ? Form the equations. (a) 3x3- 6x + 2 = 0. Solution : We wish to bring into every term such a factor that all the resulting coefficients are divisible by 3. Let A: = 3. Supply the lacking term, 3x3 + 0x2 -6x + 2 = 0. By rule, 3x3 + 3 • 0x2 - 9 • 6x + 27 • 2 = 0. • Dividing by 3, x3 - 18 x + 18 = 0. 186 ADVANCED ALGEBRA x2 (b) X3 + - - 1 = 0. (C) X8 - i = 0. (d) «» + 7x2 + -x + - = 0, 62 6» (e) x4 + ^ x2 + 1 = 0. (f) 2x8 - 3x2 -x + 4 = 0. (h) x*-6x3-2x2+ 1 =0. (g) 3x4-3x2-4x + l = 0. (i) 16x4 - 24x3 + 8x2 -2x + 1 = 0. 6. Form equations whose roots are the negatives of the roots of the fol- lowing equations. (a) x8-4x + 6 = 0. Solution : Supply the lacking term, x3+ 0x2-4x + 6 = 0. Changing signs we obtain by Corollary II x3_0x2-4x- 6 = 0, or x3 - 4 X - 6 = 0. (b) X' - 2x2 - 4x = 0. (c) x* - 3 x2 + 1 = 0. (d) X* - 2x3 + x2 4- 2x - 1 = 0. (e) x3 + 3x2 + 7x - 13 = 0. 7. What effect does changing the sign of every term of the member involving x have on the graph of an equation? 8. What is the graphical interpretation of the transformation which changes the signs of the roots of an equation, that is, what relation does the graph of the equation before transformation bear to the graph of the equation after transformation (a) when the degree is an even number, (b) when the degree is an odd number? 9. If 4x4 - 16x3 - 86x2 + 4x + 21 = has as two roots - ^ and - 3, what are the roots of 4x4 + 16x3 - 85x2 - 4x + 21 = 0? 10. If a root of xs - 11 x2 + 36 x - 36 = is 2, what are the roots of x»+ 11x2+ 36x + 36 = 0? 176. Descartes' rule of signs. A pair of successive like signs in an equation is called a continuation of sign. A pair of successive unlike signs is called a change of sign. In the equation 2x4_3a;8 + 2x2 + 2a;-3 = (1) are one continuation of sign and three changes of sign. This may be seen more clearly by writing merely the signs, + — + + —. Let US now inquire what effect if any is noted on the number of changes of sign in an equation if the equation is multiplied by THEORY OF EQUATIONS 187 a factor of the form x — a when a is positive, that is, when the number of positive roots of the equation is increased by one. Let us multiply equation (1) by a; — 2. We have then x-2 2x^ -3x*-{-2x^-{-2x''-Sx -4:X^+6x^-4:X^-4:X-\-6 2x' - 7 x^ -{- Sx^ - 2x^ - 7 x -{- 6 In this expression the succession of signs is + — + — — +, in which there are four changes of sign, that is, one more change of sign than in (1). If an increase in the number of positive roots always brings about at least an equal increase in the number of changes of sign, there must be at least as many changes of sign in an equation as there are positive roots. This is the fact, as we now prove. Descartes' rule of signs. An equation f(x) = has no more real positive roots than f(x) has changes of sign. Illustration. In the equation of degree one £c — 2 = there is one change of sign and one jjositive root. In the case of a linear equation there is no possibility of more than one change of sign. In the quadratic equation x'^-\-2x-\-l = Q there is no change of sign, and also no positive root since for positive values of x the expression x^ -f 2 cc + 1 is always positive and hence never zero. In the equation ic^ + 2ic — 3 = we have one change of sign, and one positive root, + 1. We shall prove this general rule by complete induction. First. W^e have just seen that the rule holds for an equation of degree one. Second. We assume that the rule holds for an equation of degree m, and prove that its validity for an equation of degree m -\-l follows. We shall show that if we multiply an equation of degree m hj x — a, where a is positive, thus forming an equa- tion of degree m + 1, the number of changes of sign in the new equation always exceeds the number of changes of sign in the 188 ADVANCED ALGEBRA original equation by at least one. That is, the number of chanj of sign increases at least as rapidly as the increase in the number of positive roots when such a multiplication is made. Let/(£c) = represent any particular equation of the nth. degree. The first sign of f(x) is always +. The remaining signs occur in successive groups of + or — signs which may contain only one sign each. If any term is lacking, its sign is taken to be the same as an adjacent sign. Thus the most general way in which the signs of /(x) may occur is represented in the following table, in which the dots represent an indefinite number of signs. The multiplication of f(x) by a; — a is represented schematically, onl^P"-^ the signs being given. ^ fix) X — a All + signs + •••• + All - signs All + signs All - signs Further groups + •••• + All - signs + - xf{x) - ocfix) + + ••• + - + ••• + + + ••• + + - + ••• + + +••• + + -+ + + {x-a)f{x) + d=---± -±.-.± + ±---± -±...± + ± ± -±•••±4- The ± sign indicates that either the + or the — sign may occur according to the value of the coefficients and of a. The verti- cal lines denote where changes of sign occur in f(x). Assuming that all the ambiguous signs are taken so as to afford the least possible number of changes of sign, even then in (x — cc)f(x) there is a change of sign at each or between each pair of the vertical lines, and in addition, one to the right of all the vertical lines. Thus as we increase the number of positive roots by one the number of changes of sign increases at least by one, perhaps by more. The only possible variation that could occur in the succession of groups of signs in f(x), namely, when the last group is a group of 4- signs, does not alter the validity of the theorem. THEORY OF EQUATIONS 189 We illustrate the foregoing proof by the following particular example. Let f{x) = x^ -4:X^ - x-{-2, and let a = 2. Multiply f(x), 1 + 0-4 + 0-1 + 2 4 changes by x-2, 1-2 xf(x), 1 + 0-4 + 0-1 + 2 -2f(x), -2-0 + 8-0 + 2-4 (x~2)f(x), 1-2-4 + 8-1 + 4-4 5 changes 177. Negative roots. Since /(—a:) has roots opposite in sign to those of f(x) (p. 185), we can state Descartes' rule of signs for negative roots. f(x) has nc more negative roots than there are changes in sign in /(— x). If by Descartes' rule it appears that there cannot be more than a positive roots and b negative roots, and if a -{- b <n, the degree of the equation, then there must be imaginary roots, at least n — (a -{- b) in number. EXERCISES 1. Prove Descartes' rule of signs for x^ + bx + c = directly from the expression for b and c in terms of the roots (see § 115). 2. Find the maximum number of positive and negative roots and any possible information about imaginary roots in the following equations. (a) x3 + 2 x2 + 1 = 0. Solution : Writing signs of f{x), + + +, there is no change, hence no positive root. Writing signs of /(— £c), — + + , there is one change, hence no more than one negative root. Since there can be only one real root there must be two imaginary roots. (b) x3 + 1 = 0. (c) X* - 2 = 0. (d) x3 - X + 1 = 0. (e) x6 - X + 1 = 0. (f ) x4 + X + 1 = 0. (g) x6 + x2 + 1 = 0. (h) x3-6x2+4x-l = 0. (i) x5-2x4-3x3+4x2+x + l = 0. (i) x6 + 2x* - 6x3 - 4x2 + X - 1 = 0. 190 ADVANCED ALGEBRA 178. Integral roots. In finding the rational roots of an equa- tion we make use of the following Theorem. If the equation ^ + «i^"-'+---+a„ = ^ (1) {where the oUs are integers) has any rational root, such root must he an integer. Suppose - be a fraction reduced to its lowest terms which satisfies the equation. Then ^ + 2l£r! + ... + „„ = is an identity. Then clearing of fractions and transposing, f = --q {ciip"~^ H h a„!?"~0- Thus some factor of g' is a factor of ^^ that is, oip (p. 52), which contradicts the hypothesis that - is reduced to its lowest terms. Thus all the rational roots of the equation are integers, which as we know (§ 170) are factors of a„. 179. Rational roots. If we seek the rational roots of aox"" H [- a„ = 0, where a^ =^ 1, we can multiply the roots by a properly chosen constant (§ 175) and obtain an equation of form (1) above whose integral roots may easily be found by synthetic division. Example. What rational roots, if any, has 8x»+lla;-14 = 0? (1) Multiply the roots by 8, 3 a^ + 99 x - 378 = 0. Divide by 3, x^ + 33 x - 126 = 0. (2) Since by Descartes' rule of signs equation (1) has only one positive root and no negative root, we do not need to carry the table further than to test for a positive root. THEOKY OF EQUATIONS 191 y Form a table of values for equation (2) by synthetic division. ~^ We need only to try the factors of 126 (§ 170). _ g2 Thus (2) has the root 3. Hence the original equation (1) has Q the root 3^3 = 1. KuLE. To find all the rational roots of an equation, trans- form the equation so that the first coefficient is + 1. Find the maxiymum number of positive and negative roots by Descartes' rule of signs. Find the integral roots of this equation by trial, and the roots of the. original equation by dividing the integral roots found by the constant by which the roots were multiplied. By the Theorem § 178 we are assured that all the rational roots can be found in this way. EXERCISES Find all the rational roots of the following equations. 1. 4x8 = 27(x + l). 2. 15jc8 + 13a;2-2 = 0. 3. 4a;3-6x-6 = 0. 4. x^- 2f x2 + 2|x - 1 = 0. 5. 4x3 - 8x2 - X + 2 = 0. 6. 3x* - 8x8 _ ^q^2 + 25 = 0. 7. 4x8-4x2+ x-6 = 0. 8. 3x8+ 13x2+ llx - 14 = 0. 9. 4x8 + 16x2 -9x- 36 = 0. 10. 2x3 - 21x2 + 74x - 85 = 0. 11. 6x8 - 47x2 + 71X + 70 = 0. 12. 12x8 - 52x2+ 23x + 42 = 0. 13. 6x8-29x2+ 53x- 45 = 0. 14. 6x*- x8- 8x2- 14x + 12 = 0. 15. 27x3+ 63x2+ 30x- 8 = 0. 16. 2x*-13x8 + 16x2- 9x +20 = 0. 17. 3x8 - 26x2 +52x- 24 = 0. 18. Ox* - x3 - 49x2+ 55x - 50 = 0. 19. 18x8+ 81x2+ 121x + 60 = 0. 20. 12x4 + 6x8 - 24x2- 9x + 9 = 0. 21. 10x4 + 18x8 - 16x2 + 8x- 20 = 0. 22. 9x4 + 15x8 - 143x2 +41X + 30 = 0. 23. 36x4 - 72x8 - 31 x2 + 67 x + 30 = 0. 24. 24x4- 108x8+ 324x2 -240X + 60 = 0. 180. Diminishing the roots of an equation. In the preceding sections we have solved completely the problem of finding the rational roots of an equation. We now pass to the problem of V 192 ADVANCED ALGEBRA finding the approximate values of the irrational roots of an eqnar tion. In carrying out the process that we shall develop it is desirable to form an equation whose roots are equal respectively to the roots of the original equation each diminished by a constant. Let fix) = aox"" + a^x''-'^ -\ h «« = 0, (1) whose roots are ^i, a^, • ••, a^. Let a be any constant. We seek an equation whose roots are a^ — a, a^ — a, - • • , a^ — a. ^~^^^v_ If we let a stand for any one of the roots of (1), since f(a) = (p. 33), we see that /(« + a) = is satisfied by a — a, that is, f{a-a-\-a)= f(a) = 0. Thus to form the desired equation replace x by z -\- a. We obtain f(x) = f(z + a) = ao(z + a)» ■}- a,{z -{- af-^ + . . . + a„ = 0. Developing each term by the binomial theorem and collecting like powers of z, we get an equation of the form fix) = Fiz) = A,z- + A^z-^ + . . . + ^^ = 0, (2) where the ^'s involve the a's and d. This is the equation desired. We now seek a convenient method of finding the values of the coefficients ^o? ^i? ^2> •••> ^n when a^^ %, a^^ •••, a„ are given numerically. Now A^ is the remainder from the division of Fiz) by z. But since Fiz) =fix) and z = x — a, the remainder from dividing Fiz) by z is identical with' the remainder from dividing fix) hj x — a. Thus A^ is the remainder from dividing fix) by X — a. Furthermore, since A^_y^ is the remainder from dividing —^^ by z, it is also the remainder from dividing ^^-^^ by a; — a. The process may be continued for finding the other ^'s. We may then diminish the roots of an equation by a as follows : EuLE. The constant term of the new equation is the remaivr- der from dividing fix) hy x — a. i THEORY OF EQUATIONS 193 The coefficient of z in the new equation is the remainder from dividing the quotient just obtained hy x — a. The coefficients of the higher powers of z are the remainders from dividing the successive quotients obtained hy x — a. Example. Form the equation whose roots are 2 less than the roots of x4 _ 2 x3 - 4 x2 4- x - 1 = 0. The divisions required by the rule we carry out synthetically (p. 169). 1-2-4+1- 1[2 + 2 +0 -8 -14 -15 1 + + 2 -4 -7 + 4 +0 1 + 2 +0 + 2 +8 -7 1 + 4 + 2 + 8 1+6 The desired equation is ^4 + 6x3 + 8x2 7x-15 = 0. 181. Graphical interpretation of decreasing roots. If an equa- tion has roots a units less than those of another equation, if a is positive its intersections with the X axis or with any line parallel to the X axis are a units to the left of the corresponding inter- sections of the first equation. It is, in fact, the same curve, except- ing that the Y axis is moved a units to the right. If a is negative, the Y axis is moved to the left. EXERCISES Plot, decrease the roots by a units, and plot the new axes. 1. x4-3x3 -2x-3 = 0. a = 3. Solution: 1-3- 0- 2-3[3 (1) + 3-0 - -6 1_0 - - 2 + 3 + 9+27 -9 1 + 3 + 9 + 3 + 18 + 25 1 + 6 + 3 1 + 9 + 27 194 ADVANCED ALGEBRA Thus the required equation is 0. (2) X y - 3 : 1 - 7 2 -16 3 - 9 -1 + 3 3. x3 _ 8 = 0. a = 1.4. 5. x3 + 4x-8 = 0. a = 3. 7. x3 + 2 X + 6 = 0. a = -\. 9. «8-2x2 + 8x-7 = 0. a = 2. 11. x4-3x2 + 2x 12. X 13. 2x3-6x2 + 4x -3 = 14. X* + 6x3 + 10x2 + x - 1 = In the figure one square on the Faxis repre- sents two units of y, and two squares on the X axis represent one unit of x. 2. x*-16 = 0. a = 2. 4. x4- 2x2 + 1 = 0. a = .2. 6. x3-4x2-2 = 0. a = .5. 8. x3 + 4 x2 + X - 6 = 0. a = - .4. 10. x3-3x2 + x-l = 0. a = -.3. 2 = 0. a = -2. 15x2 + 7x + 125 = 0. a = 5. - a = -3. a = -l. 182. Location principle. If when plotting an equation y =f(x) the value x = a gives the corresponding value of y positive and equal to c, while the value x = b gives the corresponding value of y negative, say equal to — d, then the point on the curve x = a, y = c is above the X axis, and the point on the curve x = byy = — -^ is below the X axis. If our curve is unbroken, '^ must then cross the X axis at least once between the values X — a and x = h^ and hence the equation must have a root between those values of x. The shorter we can determine this interval a to b the more accurately we can find the root of the equation. This property of unbrokenness or continuity of the graph of ?/ = ^0^" 4- aix"~^ + • • • + a„ we assume. We assume then the following THEORY OF EQUATIONS 195 Location principle. When for two real unequal values of x, x = a and x = h, the value of y —f{x) has opposite signs, the equation f{x) = has a real root between a and h. Illustration. The equation f(x) =ic^-f3cc — 5 = has a root between 1 and 2. Since /(I) = - 1, /(2) = 9. 183. Approximate calculation of roots by Horner's method. We are now in a position to compute to any required degree of accuracy the real roots of an equation. Consider for example the equation ^s + 3,_20 = 0. (1) Form the table of values for plotting the equation ic8 + 3 a: - 20 = ?/. By the location theorem we find that a root is between + 2 and + 3. To find more precisely the position of the root we might estimate from the graph the position of the root and substitute say 2.3, 2.4, and so 3+16 on, until we found two values be- tween which the root lies. We can gain the same result with much _ i _ 24 less computation if we first dimin- ish the roots of the equation so that the origin is at the less of the two integral values between which we know the root lies. Here we decrease the roots of (1) by 2, 1 4- + 3 - 20[2 y -20 -16 + 2 + 4+14 1 + 2 + 7 + 2 -h 8 - 6 1 + 4 + 2 + 15 1+6 The equation whose roots are decreased by 2 is a:» + 6ic2 + 15a;^6 = 0. (2) 196 ADVANCED ALGEBRA We know that (2) has a root between and 1, since equation (1) has a root between 2 and 3. From the graph we can estimate the position of the root. Having made an estimate, say .3, it is neces- sary to verify the estimate and determine by synthetic divi^io^ precisely between which tenths the root lies. Thus, trying .3, we" ^^*^^^ 1 + 6.0 + 15.00 - 6.000[_^ + 0.3+ 1.89 + 5.067 1 + 6.3 + 16.89 - 0.933 () which shows that for x = .3 the curve is below the X axis, hence the root is greater than .3. But we are not justified in assuming that the root is between .3 and .4 until we have substituted .4 for X. This we proceed to do. 1 + 6.0 + 15.00- 6.000 [^ + 0.4+ 2.56 + 7.024 1 + 6.4 + 17.56 + 1.024 Since the value of y is positive for x = .4, the location principle shows that (2) has a root between .3 and .4, that is, (1) has a root between 2.3 and 2.4. To find the root correct to two decimal places, move the origin up to the lesser of the two numbers between which the root is now known to lie. The new equation will have a root between and .1. This process is performed as follows : 1 + 6.0 + 15.00 - 6.000[^- + 0.3 + 1.89 + 5.067 1 + 6.3 + 16.89 + 0.3 + 1.98 1 + 6.6 + 0.3 0.933 + 18.87 1 + 6.9 Thus the new equation is x^ + 6.9 x^ + 18.87 X f .933 = 0. (3) This equation has a root between and .1. We can find an approximate value of the hundredths place of the root by solving THEORY OF EQUATIONS 197 the linear equation 18.87 x — .933 = 0, obtained from (3) by drop- ping all but the term in x and the constant term. This suggestion must be verified by synthetic division to deter- mine between what hundredths a root of (3) actually lies. 1 + 6.90 + 18.870 - 0.9330|.04 + 0.04 + 0.277 + 0.7658 6.94 + 19.147 - 0.1672 Thus the curve is below the X axis at a; = .04 and hence the root is greater than .04. We must not assume that the root is between .04 and .05 without determining that the curve is above the X axis at x = .05. 1 + 6.90 -f 18.870 - 0.9330[^ + 0.05+ 0.347 + 0.9608 6.95 + 19.217 + 0.0278 Thus the curve is above the X axis ai x = .05. By the location principle (3) has a root between .04 and .05, that is, (1) has a root between 2.34 and 2.35. We say that the root 2.34 is correct to two decimal places. If a greater degree of precision is desired, the process may be continued and the root found correct to any required number of decimal places. The foregoing process affords the following EuLE. Plot the equation. Apply the location principle to deter- mine between what consecutive positive integral values a root lies. Decrease the roots of the equation hy the lesser of the two integral values between which the root lies. Estinfiate from the plot the nearest tenth to which the root of the new equation lies, and determine hy synthetic division pre- cisely the successive tenths between which the root lies. Decrease the roots of this equation by the lesser of the two tenths between ivhich the root lies, and estimate the root to the nearest hundredth by solving the last two terms as a linear equation. 198 ADVANCED ALGEBRA Determine hy synthetic division precisely the hundredths inter- val in which the root must lie. ^ — ^ Proceed similarly to find the root correct to as many 'places ds may he desired. The sum of the integral, tenths, and hundredths values next less than the root in the various processes is the approximate value of the root. To find the negative roots of an equation f(x) = 0, find the positive roots of /(— x) = and change their signs. When all the roots are real a check to the accuracy of the com- putation may be found by adding the roots together. The result should be the coefficient of the second term. EXERCISES Eind the values of the real roots of the following equations correct to three decimal places. 1, x3 + 4x2 + aj + l = 0. (1) Solution : Since by Descartes' rule of signs there are no posi- tive roots, we form the equation /( — a;) = and seek its positive root. Thus a;3-4x2 + aj-l = 0. (2) Plot the equation (2) set equal to y. In the figure two squares are taken as the unit of x. There is a root of this equa- tion between 3 and 4. Yj < ( X i ^ / \ 1 > 2 k ,f 3 \ 4 V J \ / ^ > y -1 -3 -7 -7 + 3 Decrease the roots of (2) by 3, 1-4+1 + 3-3 1[3 6 1-1 -2 + 3 +6 1 + 2 + 3 1 + 5 + 4 THEORY OF EQUATIONS 199 The equation is x^ + 5x^ + Ax — 7 = 0. (3) From the plot we estimate the root of (3) at .8. Verify, 1 + 5.0 + 4.00 - 7.000 [^ + 0.8 + 4.64 + 6.912 + 5.8 + 8.64 - 0.088 1+ 5.0 + 4.00- 7.000 [^ + 0.9 + 5.31 + 8.379 + 5.9+9.31 + 1.379 Thus the root is determined between .8 and .9. \ Decrease the roots of (3) by .8, 1+5.0 + 4.00 -7.000^8 + 0.8 + 4.64 + 6.912 1+5.8 + 8.64 + 0.8 + 5.28 .088 1+6.6 + 0.8 + 13.92 1+7.4 The equation is x^ + 7.4 x^ + 13.92 x - .088 = 0. (4) 088 Estimate the root of (4) at x = ~ = .006. ^ ' 13.92 Verify, 1 + 7.400 + 13.920000 - .088000 [.006 + 0.006 + 00.044436 + .083784 + 7.406 + 13.964 - .004216 1 + 7 .400 + 13.920000 - .088000 [.007 +0 .007 + 00.051849 + .097804 + 7.407 + 13.972 + .009804 Thus the root of the equation (1) correct to three decimal places is - 3.80(> 2. x3 - 4 = 0. 3. x* - 3 = 0. 4. X3 + X = 20. 5. 3X4 - 5X3 zz: 3I. 6. x3 + x2 = 100. 7. x3 - X - 33 = 0. 8. x4 + X - 100 = 0. 9. x3 - 8x - 24 = 0. 10. x4 - 4x3 + 12 = 0. 11. X* + x2 + X = 111. 12. x3 - x2 + X - 44 = 0. 13. x3 + lOx - 13 = 0. 14. x8 + 3 x2 - 2 X - 1 = 0. 15. x3 + x2 + X - 99 = 0. 16. x3 - 9x2 - 2 X + 101 = 0. 17. X* + x3 + x2 - 88 = 0. 18. X* - 12x3 - 16x + 41 = 0. 19. x3 - 6x2 + 5x + 11 = q. 20. x3 - 10x2 + 35x+ 50 = 0. 21. 2x4-4x3 + 3x2-1 = 0. 22. 3x4 - 2x3 - 21x2 -4x + 11 = 0. 23. 9x3 - 45x2 + 34x + 37 = 0. 200 ADVANCED ALGEBRA 184. Roots nearly equal. Suppose we wish to find the positije roots, if any exist, of ) x^-{-17x^-A6x-i-29 = 0. (1) y 4-29 + 1 + 13 + 71 By Descartes* rule of signs we see that there can be only two positive roots. We obtain the adjacent table of values. From the plot that these values indicate we cannot tell whether any real root exists between 1 and 2, but if it does exist the plot indicates that it is nearer 1 than 2. Decrease the roots of (1) by 1, 1 + 17 -46 +29[1 + 1+18-28 1 + 18 - 28 + 1 + 1+19 1 + 19 - 9 + 1 1+20 The new equation is x^-\-20x^- -9x- f 1 = 0. (2) Estimate the root of (2) at .2 and carry the origin up to .2 and also up to .3. 1 + 20.0 - 9.00 + l.OOOj^ + 0.2 + 4.04 - 0.992 1 4- 20.0 - 9.00 + 1.000[J + 0.3 + 6.09 - 0.873 1 + 20.2 -4.96 + 0.2 + 4.08 1 + 20.4 + 0.2 + 0.008 1 + 20.3 - 2.91 + 0.3 + 6.18 0.88 1 + 20.6 + 0.3 + 0.127 + 3.27 1 + 20.6 1 + 20.9 By Descartes' rule of signs on the numbers obtained by moving the origin to .3, it is seen that there are no positive roots of (2) greater than .3, while the rule would indicate that there might be roots greater than .2. We consider the equation ic^ + 20.6 x^ - 0.88 X + 0.008 = 0. (3) THEORY OF EQUATIONS 201 Estimate the roots of (3) at _ 0.008 ^ ~ 0.88 = .009.=* Verify, 1 + 20.60 - 0.880 -h 0.00800[.01 + 0.01 + 0.206-0.00674 1 -f 20.61 - 0.674 + 0.00126 1 + 20.60 - 0.880 + 0.00800|.02 + 0.02 + 0.412 - 0.00936 1 + 20.62 - 0.468 - 0.00136 This determines a root of (3) between .01 and .02. 1 + 20.60 - 0.880 4- 0.00800|.03 + 0.03 + 0.619-0.00483 1 + 20.63 - 0.161 + 0.00317 This determines another root of (3) between .02 and .03. Decrease the roots of (3) by .01, 1 + 20.60 - 0.880 + 0.00800|.01 + 0.01 + 0.206 - 0.00674 1 + 20.61 - 0.674 + 0.01 +0.206 + 0.00126 1 + 20.62 + 0.01 - 0.468 1 + 20.63 The new equation is x^ + 20.63 x^ - 0.468 x + 0.00126 = 0. (4) Estimate the root of (4) at ic = ' ,^^ = .002.=* ^ ^ 0.468 Verify, 1 + 20.630 - 0.468 + 0.00126|.002 + 0.002 + 0.041 - 0.00085 1 + 20.632 - 0.427 + 0.00041 1 + 20.630 - 0.468 + 0.00126|.003 + 0.003 + 0.062-0.00122 1 + 20.633 - 0.406 + 0.00004 ♦ We observe that in these two cases the estimated values of the roots are shown by the verification to be inaccurate. This should insure great care in making the verifi' cation. Th^ estimated values should nev^ b^ assumed to be accurate without verification 202 ADVANCED ALGEBRA This indicates that a root is between .003 and .004. Verify, 1 + 20.630 - 0.468 + 0.00126|.004 + 0.004 + 0.083-0.00154 1 + 20.634 - 0.385 - 0.00028 This determines a root of (3) between .003 and .004. Thus one root of (1) correct to three decimal places is 1.213. The other root could be found similarly to be 1.229. EXERCISES Find all the real roots of the following equations correct to three decimal places. 1. x^-1x-^1 = 0. 2. 7x3 -8«2_i4x + 16 = 0. 3. 2ic5_ 4x3-3x2 + 6 = 0. 4. 4x4- 5x3 -8x + 10 = 0. 5. 3x3 - 10x2 - 33x + 110 = 0. CHAPTEE XVIII DETERMINANTS 185. Solution of two linear equations. We have already treated the solution of linear equations in two variables and stated (p. 47) the method of solving three or more linear equa- tions in three or more variables. This latter process is rather laborious and can be very much abridged and also developed more symmetrically by the considerations of the present chapter. Let us solve the equations a^x + b^ij = Ci, (1) a^x + b^y = Ca. (2) Multiply (1) by b^ and (2) by b^, and we obtain ai^2^ 4- bib^i/ = b2Ci ^. v Subtracting, we get (aib^ — a^bi) x = b^c^ — biC^, or if aA - ^2^1 ^ 0, ^^h^LIzh^. (3) ai/>2 — ^2^1 Similarly, we obtain y = —^ ^' (4) We note that the denominators of the expressions for x and y are the same. This denominator we will denote symbolically by the following notation : ^, aibz — a^bi = , • _ ^2 t*2 The symbol in the right-hand member is called a determinant. Since there are two rows and two columns, this determinant is said to be of the second order. The left-hand member of the equa- tion is called the development of the determinant. The symbols Ui, bi, a2, ^2 ^^6 called elements of the determinant, while the ele- ments «! and &2 are said to comprise its principal diagonal. 203 (6). 204 ADVANCED ALGEBRA EuLE. The development of any determinant of the seconc order is obtained by subtracting from the product of the ele^ ments on the principal diagonal the product of the elements on the other diagonal. Thus Zi xyi - ziy ; 12 3 4 5 10 - 12 = - 2. Evidently each term of the development contains one and only one element of each row and each column, that is, for instance, the letter a and the subscript 1 appear in each term of (5) once and only once. We can now rewrite the solution (3) and (4) of equations (1) and (2) in determinant form : X = Cl h ai Ci C2 h ; y = ^2 ^2 ai b. a^ bi a^ b. a^ b^ (6) It is noted that the numerator of the expression for x is formed from the denominator by replacing the column which contains the coefficients of x, a^ and a^,, by the constant terms c^ and c^. Similarly, in the numerator of the expression for y, b^ and b^ of the denominator are replaced by Cj and c^. One should keep in mind that a determinant is merely a sym- bolic form of expression for its development. In the case of determinants of the second order the introduction of the new notation is hardly necessary, as the development itself is simple ; just as we should scarcely need to introduce the exponential notation if we had to consider only the squares of numbers. It turns out, however, as we shall see, that we are able to denote by determinants with more than two rows and columns expressions with whose development it would be very laborious to deal. 186. Solution of three linear equations. Let us solve the equations a^x + b^y 4- Ci« = c^i, a^x 4- b^2.y + ^2^ = ^2> <iz^ + b^y + Cg^ = c?8- (1) (2) (3) DETERMINANTS 205 Eliminating y from (1) and (2) and (1) and (3), we obtain (ai^g — a^h^x H- (b^c^ — hiCc^)z = dj)^ — d^b^j (a^bi — a^b^x + {c^b^ — b^c^z — d^b^ — d-J)^. Eliminating z, [iSHh — «2^i) (^3^1 — h(^i) — (ctsbi — a^bs) (b^Ci — ^iC2)]a; = (dibi — d^bi) (c^bi — bsCi) — (d^bi — dj)^) (b^Ci — b^c^} Developing, canceling, and dividing by bi, we obtain _ d^b^Cs + d^bsCi + ^3^1 ^2 — dib^c^ — dsb^c^ — d^b^c^ aib^c^ + a^b^c^ + a^b^c^ — a^b^c^ — a^b^c^ — a^Jy^c^ ^ Following the analogy of tbe last section, we write the denomi- nator ai^2^3 + «2^8^1 + «8^1^2 ~" %^3<'2 "~ «3^2Cl ~ «2^1^3 = «1 l\ Cl ^2 K ^2 ^3 bs Cs (5) The right-hand member of this equation we call a determinant of the third order, and the left-hand member its development. As in the determinant of the second order, the elements a^, b^, Cg comprise the principal diagonal; each term of the development contains one and only one element of each row and each column, and all possible terms so constructed are included in the develop- ment. The signs of the terms of the determinant of the third order may be kept in mind by the following device. ' Rewrite the first and second columns to the right of the deter- minant as follows : «! ^1 G^ <Zi &i % ^2 ^2 ^2 h «3 ^8 ^3 % h The positive terms are found on the diagonals running down from left to right, the negative terms on the diagonals running up from left to right. The numerator of the fraction (4) expressed in determinant notation is «! bi Cj dz b^ Cg dn b^ Cg 206 ADVANCED ALGEBRA We can find similarly the values of y and z that satisfy equ£ tions (1), (2), and (3). The solutions of the equations in determi- nant form are as follows : d. ^1 ^2 d. b. d. h ^3 a^ K ^1 ^2 h <^2 ^3 h Oz y = ^1 d. Ci ^2 d. C2 ^3 d. Cz «1 h ^1 ^2 h ^2 ^3 h Cz «i *i d. aj *s d. as Js d> a, h h a. h <'2 as h Cs (6) The same principle observed on p. 204 for forming the deter- minants in the numerators of the expressions for x and y may be followed here. The determinant in the numerator of the expres- sion for ic, 2/, or z is found from the denominator by replacing the column that contains the coefficients of the variable in question by a column consisting of constant terms. Thus in the numerator of z we find the column d-^, d^^ d^ replacing the column Cj, Cg, Cg of the denominator. EXERCISES 1. Find the value of the following determinants. 3 2 1 (a) 4 6 2 1 1 • 3 2 1 Solution : 4 6 2 1 1 4 3 1 (b) 1 2 6 1 1 3 4 1 (d) 2 4 1 6 a X y (f) b c c b • c b (h) -c a -b -a o| = 18 + 4 + 0-6-8-0 = 8. 2 1 (c) 2 1 1 2 2 1 1 (e) 6 3 3 4 2 3 a b c (g) b c a c a b c b (i) c a b a DETERMINANTS 207 2. Solve the following equations by determinants. (a) 2x + 32/ = 4, x-2y = l. Solution : Using the expressions (6), p. 206, we obtain -8 -3 _ n -4-3~y 4 3 1 -2 2 3 1 -2 y = 2 4 1 1 2 3 1 -2 (e) 7 y = 12, 7 a: + 2/ = 11. 2x+5y=l, 2x + 72/:-l, a: + 4y = 2, ^ ' 7a;+6?/ = 2. ^ -* Sx-9y = 2. ^ ' 2a;-3?/ = 3. Solve the following equations by determinants. a: + y + 2 = 2, (a) X + 32/- 4 = 0, y - 2 2; = 6. Solution : Rearranging so that terms in the same variable are in a column, and supplying the zero coefficients, we get x+ 2/4- 2 = 2, a: + 32/ + 02 = 4, Ox+ y-2z = 6. By (6), p. 206, x 2 1 1| 4 3 6 1 -2 J 1 1 1 3 1 -2 -12+0 + 4-18 + 8 + 18 1 I /-^6 + + l 0-0.+ 2 = 6, 1 2 1 1 4 6 -2 -8+0+6-0+4+0 1 1 1 -3 1 3 1 -2 1 2 3 4 1 6 1 1 1 13 1-2 Check: 6 + (-|) + (--L0) = ,18 + + 2-0-6-4 -3 6-4 = 2. 10 10 208 \ AD 2ic + 3y = 12, (b) 3a; + 2:2 = 1 Sy + 4:Z x + y -z (d) x + z~ V + z- x + y + (f) 3a:-2z = 5 y — 4 z = 0. X + 2/ + 2 = 9, (h) x + 2y + 3z = 14, jc + 3y + 62; = 20. .2x + .3y + .4z = 29, (j) .3x + .4y + . 521=38, .4a; + .5y + .6z = 51. ED AMe^EBRA 7 ^lx-iy = 0, Ac) ix-iz = l, 2; - i y = 2. a + 2y = 7, ) 7x+92; = 29, y+8z = 17. X + y + 2 2; = 34, (g) X + 2 y + 2: = 33, 2 X + y + 2: = 32. ax + 6y — C2; = 2 a6, (i) 6y + C2; — ax = 2 6c, C2; + ax — 6y = 2 ac. 3x + 2y + 32: = 110, (k) 5 X + y - 4 2; = 0, 2x-3y + 2; = 0. 187. Inversion. In order to find the development of determi- nants with more than three rows and columns, the idea of an inver- sion is necessary. If in a series of positive integers a greater integer precedes a less, there is said to be an inversion. Thus in the series 12 3 4 there is no inversion, but in the series 12 4 3 there is one inversion, since 4 precedes 3. In 1 4 2 3 there are two inversions, as 4 precedes both 2 and 3 ; while in 1 4 3 2 there are three inversions, since 4 precedes 2 and 3, and also 3 precedes 2. 188. Development of the determinant. In the development of the determinant of order three we have %^2^3 + «2^3Cl + ^S^l^a — 0^8^2Cl — ^2^1^8 " «1^3^2- (1) the order of lettejps in each term the same as their in the principal diagonal (as we have done in the development above), it is observed that the subscripts in the various terms take on all possible permutations of the three digits 1, 2, and 3. The permutations that occur in the positive terms are 12 3, 2 3 1, 3 1 2, in which occur respectively 0, 2, and 2 inversions. The permutations that occur in the subscripts of the negative terms are 3 2 1, 2 1 3, 1 3 2, in which occur respectively 3, 1, and 1 inversions. DETERMINANTS 209 Thus in the subscripts of the positive terms an even number of inversions occur, while in the subscripts of the negative terms an odd number of inversions occur. This means of determining the sign of a term of the development we shall assume in general. When we have a determinant with 71 rows and columns it is called a determinant of the nth order. The development of such a determinant is defined by the following KuLE. The development of a determinant of the nth order is equal to the algebraic sum of the terms consisting of letters fol- lowing each other in the same order in which they are found in the principal diagonal hut in which the subscripts take on all possible permutations. A term has the positive or the negative sign according as there is an even or an odd number of inver- sions in the subscripts. This means of finding the development of a determinant is use- ful in practice only when the elements of the determinant are letters with subscripts such as in (2) below. When the elements are numbers we shall find the value of the determinant by a more convenient method. In this statement it is assumed that the number of inversions in the subscripts of the principal diagonal is zero. If this number of inversions is not zero, the sign of any term is + or — accord- ing as the number of inversions in its subscripts differs from the number in the subscripts of the principal diagonal by an even or an odd number. Since each term contains every letter a, b, ••-jk and also every index 1, 2, • • • , ri, one element of each row and column occurs in each term. In the determinant of the fourth order tti 61 Ci di a^ 62 C2 dz 03 &3 C3 ^i a^ &4 C4 d^ the terms a^h^Cidi and a^h^c^di, for instance, have the minus sign, as 2 4 1 3 haa three inversions and 4 2 3 1 has five inversions; while the terms aih^c^d^ and a^b^Cidi have two and six inversions respectively and hence have the positive sign. 210 ADVANCED ALGEBRA 189. Number of terms. We apply the theorem of permutar tions to prove the following Theorem. A determinant of the nth order has n ! terms in its development Since the number of terms is the same as the number of per- mutations of the n indices taken all at a time, the theorem follows immediately from the corollary on p. 145. 190. Development by minors. In the development of the determinant of order three, p. 208, we may combine the terms as follows : <^l (pi^^Z — h<^2) — »2 (P\<^Z — h(^l) + 0^3 (^1^2 — h<^i) 3 Cs + Cli (1) IS We observe that the coefficient of a^ is the determinant that we obtain by erasing the row and column in which aj lies. A similar fact holds for the coefficients of a^ and a^. The determi- nant obtained by erasing the row and column in which a given element lies is called the minor of that element. Thus / the minor of cig- ^^ notice that in the above development by minors (1) the sign of a given term is + or — according as the sum of the number of the row and the number of the column of the element in that term is even or odd. Thus in the first term a^ is in the first row and the first column, and since 1 -f- 1 = 2, the statement just made is verified for that case. Similarly, aj is in the first column and the second row, and since 1 + 2 = 3 is odd, the sign is minus and the law holds here. The last term is positive, which we should expect since a^ is in the first column and the third row, and 1 + 3 = 4. The proof for the general validity of this law of signs is found on p. 215. The elements of any other row or column than the first may be taken and the development given in terms of the minors with DETERMINANTS 211 — (^2 Cl + ^'2 «1 Cl — ^2 a^ ^3 dz <^Z as respect to such, elements. For instance, take the development with respect to the elements of the second row, d^ ^2 ^2 The rule of signs is the same as given above ; that is, for instance, the last term is negative, as c^ is in the third column and the second row, and 2 + 3 = 5. By generalizing these con- siderations we may find the development of a determinant by minors by the following EuLE. Write in succession the elements of any row or column, each multiplied by its minor. Give each term a + or a — sign according as the sum of the number of the row and the number of the column of the element in that term is even or odd. Develop the determinant in each term by a similar process until the value of the development can be determined directly by multiplication. That this rule for development gives the same result as the definition given in § 188 we have seen for a determinant of order three. The fact holds in general, as we shall prove (p. 215) . EXERCISES 1. In the determinant of order four on p. 209 what sign should be pre- fixed to the following terms ? (a) Ct^hzO^di. Solution: c^^hza-idi = azhcidi. In 2 3 4 1 there are three inversions. The sign should be minus. (b) a^hcsdi. (c) aibic^ds. (d) b4Cidsa2- (e) ^8610402. (f) d2aic^b3. (g) 0208^164- . 2. Develop by minors the following and find the value of the determinant 2 4 (a) Solution : 1 4 1 6 3 2 3 = 3 = 3.(6 4) |1 2(12 4) + 3 (8 - 4) = 6 - 16 + 12 = 2. 212 ADVANCED ALGEBRA 1 4 6 (p) 7 8 2 1 3 1 a b (e) d c e f • 2 1 (c) 2 1 3 4 a b (f) a b a 6 (d) 5 7 3 7 -2 3 2 1 - 1 (g) a 6 c d e • (h) Solution : Develop with respect to the elements of the first column, 2 3 3 \l 4 2 3 1 112 2 12 12 3 21 2 31 2 3 3 1 1 2 12 12 3 1 2 + 4 3 11 112 12 3 2 3 1 1 2 3 + 1 + 1 : s!)-('i; -2 1 2 3 1 1 1 2 1 1 1 2 1 2 1 + 1 + 2 1 2 (i) 2 6 3 i. ' = 2(-l + 2 + 0)-3(-3-2 + l) + 4(-3-l + l)-3(0-l+2) = 2 + 12 - 12 - 3 = - 1. 1 2 8 6 2 9 1 3 1 8 1 4 ax 1 1 3 3 3 4 1 1 1 1 3 3 Hint. It is always advisable to develop with respect to the row or column that has a maximum number of elements equal to zero. (k) (m) (0) (q) a b b a b a b a 2 3 4 1 2 3 3 6 2 3 1 1 2 3 2 1 X a b b X a . a b X X y X y X y X y (1) (n) (P) (r) «! 6i Ci di 62 C2 di ai a2 62 as C3 f^a ^4 63 a^ 64 C4 a gr e a / a d c b X a b c X a b c X 0, 6 c cs a c b a X DETERMINANTS 213 191. Multiplication by a constant. In this and the following sections we shall give a number of theorems on determinants which greatly facilitate their evaluation and which make a proof for the solution in terms of determinants of any number of linear equations in the same number of variables a simple matter. Theorem. If every element of a row or a column is multi- plied hy a number m, the determinant is multiplied by m. Suppose that every element of the first row of a determinant is multiplied by m. Since each term of the development contains one and only one element from the first row, every term is multi- plied by m, that is, the determinant is multiplied by m. Illustration. mbi bi ^3 mci Cg C3 = maiJg^a + fnazb^Ci -}- ma^biC^ — maib^c^, — ma^biC^ — ma^b^Cx «! a 1 a. = m bx b^ ba Ci C2 C3 • 6 4 1 Similarly, 8 3 2 = 10 4 1 23 24 2-5 = 2 192. Interchange of rows and columns. We now prove the Theorem. The value of a determinant is not changed if the columns and rows are interchanged. Take for instance the determinant of order four. «! bi Cl di ai Oa as a^ aa b. C2 d. bi b. bs b. ttg bs Cz d. Ol ^2 C3 Ci a^ b. C4 d. d. d. d. d. In each of these determinants the principal diagonal is the same, and hence the developments derived according to the statement on 214 ADVAl^CED ALGEBRA p. 209 will be the same for each determinant, since the terms wil be identical except for their order. The same reasoning is valic for any determinant. 193. Interchange of rows or columns. We now prove the Theorem. If two columns or two rows are interchanged, the sign of the determinant is changed. Again let us take for example the determinant of order four and fix our attention on the first and second rows. We must prove that «! h Ci d. a^ h. ^2 d. as h Cl d. - «! W Cl d. as h Cs ds as h Gz ds a^ h Ci d. a^ b. C4 d. In the first determinant the principal diagonal is a^bc^c^d^, while in the second the principal diagonal, a^b^^c^d^, is obtained from the principal diagonal of the first determinant by one inversion of subscripts. Hence this term is found among the negative terms of the first determinant. Since the only difference between the second determinant and the first is the interchange of the subscripts 1 and 2, evidently any term of the second is obtained from some term of the first by a single inversion. Thus if a single inversion is carried out in every term of the first determinant, we obtain the various terms of the second. But since this process changes the sign of each term of the first determinant (p. 213), we see that the second determinant equals the negative of the first. Similar reasoning may be applied to the interchange of any two consecutive rows or columns of any determinant. Consider now the effect of interchanging any two rows which are separated we will say by k intermediate rows. To bring the lower of the two rows in question to a position next below the upper one by successive interchanges of adjacent rows, we must make k such interchanges. To bring similarly the upper of the two rows to the position previously occupied by the other requires k-\-l further interchanges of adjacent rows. Hence the interchange of the two rows is equivalent to 2 A; + 1 interchanges of adjacent DETERMINANTS 215 rows, the effect of which is to change the sign of the determinant, since 2 ^ + 1 is always an odd number. 194. Identical rows or columns. This leads to the important Theokem. If a deteimiinant has two rows or two columns identical^ its value is zero. Suppose that the first and the second row of a determinant are identical. Suppose that the value of the determinant is the num- ber D. By § 193, if we interchange the first and second rows the value of the resulting determinant is — D. But since an inter- change of two identical rows does not change the determinant at all, we have j) ___ t) or 2 Z> = 0, that is, D = 0. Corollary. If any row (or column) is m times any other row (or column), the value of the determinant is zero. By § 191, the determinant may be considered as the product of m and a determinant which has two rows (or columns) identical. Hence this product equals zero. 195. Proof for development by minors. On referring to the rule on p. 211 we observe that in order to show that the develop- ment by minors is the same as the development obtained by the definition on p. 209 we must prove the two following statements. First. The coefficient of any element in the development of a determinant (apart from sign) is the minor of that element. Second. Each element times its minor should have a -\- or a — sign according as the sum of the number of the row and the column of the element is even or odd. Consider the element ^j. First. Each term that contains ai must contain every other letter than a, and the indices of these letters must take on all permutations of the numbers 2, 3, ■ • -, n. This coefficient of ai contains then by definition (p. 210) all the terms of its minor. Second. Since in each term ai is in the first place, the only inver- sions in the subscripts are those among the numbers 2, 3, • • • , ?i. 216 ADVANCED ALGEBRA Hence the sign of each term in the coefficient of ai is positive or negative according as there is an even or odd number of inversions in its subscripts. Hence our theorem is established for the ele- ment %. Consider now any element, as d^ , which occurs in the fifth row and the fourth column. Interchange adjacent rows and columns until c?5 is brought into the leading position in the principal diagonal. This requires in all seven interchanges, three to get the c?6 in the first column, and then four to get it into the first row. This changes the sign of the determinant seven times, leaving it the negative of its original value. By the reasoning just given in the case of % the coefficient of d^ (which is now in the position previously occupied by ai) in the original determi- nant would be the minor of d^, except that the signs would all be changed. Hence the term consisting of d^ times its minor has the — sign, and the theorem is proved for this case. In general, to bring a term in the ith. row and kth. column to the leading position requires i — l-{-k — l = i-\-k — 2 inter- changes of adjacent rows or columns. This involves i -\- k — 2, or what amounts to the same thing, i + k changes of sign. Hence a positive or a negative sign should be given to an element times its minor according as i + ^ is even or odd. 196. Sum of determinants. We now prove the fact that under certain conditions the sum of two determinants may be written in determinant form. The fact that the product of two determinants is always a determinant is extremely important for certain more advanced topics in mathematics, but the proof lies beyond the scope of this chapter. Theorem. If each of the elements of any row or any column con- sists of the sum of two numbers, the determinant may he written as the sum of two determinants. For example, we have to prove that «! + a'l ^1 Cl «! h Cl a\ ^1 Cl a^ + a\ h C2 = ^2 h ^2 + a'. h. Ci «» -f a's h ^3 ^8 h Cz «'s K Cs DETERMINANTS 217 Develop the first determinant by minors with, respect to the first column, where we symbolize the minors of Ui + a'j, a^ + a'2, ftg + a's by Ai, A^^ A^ respectively. We have by § 190, «! + a'l bi Ci ^2 + a'2 ^2 ^2 as + a's ^8 ^8 «! ^1 Cl a\ h Cl a^ *2 C2 + a\ h ^2 as ^8 Cs a\ h Cz = (»! -f a'O^l — (^2 + ^'2)^2 +(«8 + a'8)^8, by the distributive law, = aj^i — a^Az 4- 0.3^3 + a\Ay — a'2^2 + «'8^3> by § 190, The method of proof given is applicable to the case of any row or column of a determinant of order n. 197. Vanishing of a determinant. For the solution of systems of linear equations we shall make use of the Theorem. If in the development of a determinant in terms of the minors with respect to a certain column (or row) the ele- ments of that column (or row) are replaced hy the elements of another column (or row), the resulting development equals zero. For example, we have «! h Ci d. a? h Ci d. as h Cz d. a^ h C4 d. a-^A-i — a^A^ -p ^sA^ — a^A^j (1) where an A represents the minor of the a with the same sub- script. We have to prove that if we replace the a's, for example, by the b'Sj the result equals zero, that is, that biAi — b^A^ + ^3^3 — b^A^ — 0. (2) 218 ADVANCED ALGEBRA This expression (2) when written in determinant form evidently would have the same form as the left-hand member of (1) excej ing that the first column would consist of ^i , ^2 ? ^3 j ^4 • ^^ should then have two identical columns of the determinant, which would then equal zero (§ 194). Thus the development in (2) vanishes identically. This method of proof is perfectly general. Corollary. The value of the determinant is unchanged if the elements of any row (or column) are replaced hy the elements of that row (or column) increased hy a multiple of the elements of another row (or column). Thus, for instance. «! ^1 Ci ^2 h. ^2 = as h Cs «! -i-nb. h Ci ^2 + nh^ h ^2 ds + nh. h Cz The proof follows directly from §§ 191, 196, and the preceding theorem. 198. Evaluation by factoring. If in a determinant whose elements are literal two rows or two columns become identical on replacing a by h, then a — b i^ a factor of the development. This appears immediately from § 160. Illustration. Evaluate by factoring (1) Since two columns become identical if a is replaced by h, a by c, or h by c, then we have as a factor of the development (a-b){b-c){c-a). (2) To determine whether the signs in this product are properly chosen, that is, whether the development should contain a — b oy b — a,we note that the term bc^ is positive in the development of (1) and also positive in the expansion of (2). Evidently there is no factor of (1) not included in (2). 1 1 1 a b c a^ b' c^ DETERMINAKTS 219 199. Practical directions for evaluating determinants. In j&nding the value of a numerical determinant the object is to reduce it to one in which as many as possible of the elements of some row or column are zero. One should ask one's self the following questions on attempting to evaluate a determinant: First. Is any row (or column) equal to any other row (or column) ? If so, apply § 191 for the case m = 0. Second. Are the elements of any row (or column) multiples of any other row (or column) ? If so, apply § 191. Third. If we add (or subtract) a multiple of the elements of one row (or column) to the elements of another, will an element be zero ? If so, apply § 197. EXERCISES Evaluate the following determinants. 2 3 4 3 4 5 3 2 1 2 7 6 1 8 7 Solution : We obseive that if we subtract each element of the first column from the corresponding element of the second column, the new second column has every element 1. A similar result is obtained by subtracting the last col- umn from the third column. Thus, by § 191, 2 3 4 3 2 1 1 4 5 3 2 4 1 1 1 2 7 6 "7 1 1 1 1 8 7 1 1 = 0. 4 3 12 6 113 4 2 12 3 6 2 1 Solution : Multiplying the last column by 2 and the whole determinant by \ does not change the value of the determinant (§ 191). Thus 4 3 12 4 3 14 6 113 _ 1 6 116 4 2 12 ~ 2 4 2 14 3 6 2 1 3 6 2 2 220 ADVANCED ALGEBRA Subtracting the last column from the first column and developing, we obtam 4 3 1 4 6 1 1 6 _ 1 4 2 1 4 ~2 3 6 2 2 3 14 116 2 14 16 2 2 3 14 1 1 6 2 14 Subtracting the last row from the first row, 3 1 1 1 2 1 1 6 1 4 -,-(-2) = l. a — d a 1 b-d b 1 c — d c 1 a^ + b 2ab 2ab 1 a2 + 62 1 2a6 + 62 1 5. 7. 9. 3 3 4 2 1 1 2 1 2 2 3 1 2 1 3 2 a; y 2 X y 2 y 2 « 2 y a; a 1 & 1 1 c 1 2 d 1 3 3 10. 4 6 8 3 1 1 2 1 2 3 4 1 2 1 3 4 110 1 10 11 111 1111 1 1 1 1 1 1 1 1 1 1 1 1 11. a a a a a b b b a b c c a b c d 12. 2 1 1 1 1 2 1 1 1 1 2 1 1 1 1 2 13. 3 7 16 14 6 15 33 29 1 1 1 4 2 3 1 14. 8 2 1 4 16 29 2 14 16 19 3 17 33 39 8 38 15. 9 13 17 4 18 28 33 8 30 40 54 13 24 37 46 11 16. [2 14 16 18 2 4 6 8 4 3 2 1 3 7 11 16 DETERMINANTS 221 D, 200. Solution of linear equations. Suppose that we have given n linear equations in n variables. We seek a solution of the equa- tions in terms of determinants. For simplicity, let n — 4:. Given a^x + hy + Ci« + d^w =/i, (1) a^x -f h^y + c^z + d^w =/2, (2) (^3^ + hy + CsZ + d^w =/8, (3) a^x + hy + C4^ + d^w =/4. (4) The coefficients of the variables taken in the order in which they are written^ift^ be taken as forming a determinant D, which we call the determinant of the system. Thus % hi Ci di (^2 t>2 ^2 ftg ^8 ^3 ^3 ^8 a^ b^ C4 d^ Symbolize by ^1, ^3, etc., the minors of %, 63, etc., in this deter- minant. Let us solve for x. Multiply (1), (2), (3), (4) by ^1, A^, Azj A^ respectively. We obtain A^a^x + A^biy + A^c^z + A^d^w = ^j/i, A^a^x + A^b^^y + ^3^2^ + A^d^w = ^2/2, ^gaga: + A^b^y + ^3^3^ + ^af/gi^ = ^3/3, A^a^x -f- ^4^'4?/ 4- A^c^z + ^4C?4i^; = A^f^. If we add these equations, having changed the signs of the second and fourth, the coefficient of x is the determinant Z>, while the coefficients of y, z^ w are zero (§ 197). The right-hand mem- ber of the equation is the determinant D, excepting that the ele- ments of the first column are replaced by fufi, fs,f^ respectively. Hence X = /l h Cl d. A h Ca d. A h Cs d. A h ^4 d. «! h Cl d. ^2 h C2 d. ^8 h Cz d. a^ h Ci d. 222 ADVANCED ALGEBRA In a similar manner we can show that the value of any variable which satisfies the equations is given by the following EuLE. The value of one of the variables in the solution of n linear equations in n variables consists of a fraction whose denominator is the determinant of the system and whose numer- ator is the same determinant, excepting that the column which con- tains the coefficients of the given variable is replaced by a column consisting of the constant terms. When D = 0, we cannot solve the equations unless the numer- ators in the expressions for the solution also vanish. Illustration. Solve for x ax -\-2by — 1, 2by + ^cz = 2, S cz -{- 4: dw = Sj 4 dw -{- 5ax = A. E-earranging, we obtain ax + 2 by = ly 2by-^3ez =2, 3 cz -{- 4: dw = 3j -{- 4:dW = 4:. D 5 ax a 2b 2b 3c 3c 5a v" Ad = 24 4:d a 5 5 c c ^ 5a ^ ^ d 1 b 1 b 24 2 b c 1 c 3 c d 3 c d X — 4 d 4 d a b a b 24 b c b c c d c d 5a d -5b d 1 c -b 3 c d 4 d b c a c d -5b d DETERMINANTS 223 1 c -h 2 d 4 d b G a C d 5c d be ab G d 5c d - 2 bed — 4 abed 1_ 2a 201. Solution of homogeneous linear equations. The equa- tions considered in the previous section become homogeneous (p. 115) if /i =/2 =/3 =/, = 0. We have then a^x + b-^y -\- GiZ + diW = 0, azX -f bzy + G2% + dzW = 0, ^3^5 + b^y + ^3^; + cZgW; = 0, (I) These equations have evidently the solution x = y = z = w =0. This we call the zero solution. We seek the condition that the coefficients must fulfill in order that other solutions also may exist. If we carry out the method of the previous section, we observe that the determinant equals zero in the numerator of every fraction which affords the value of one of the variables (§ 191). Thus if D is not equal to zero, the only solution of the above equations is the zero solution. This gives us the following Pkinciple. a system of n linear homogeneous equations in n variables has a solution distinct from the zero solution only when the determinant of the system vanishes. Whether a solution distinct from the zero solution always exists when the determinant of the system equals zero we shall not determine, as a complete discussion of the question would be beyond the scope of this chapter. Theorem. If x^, y^, z^, w^ is a solution of equations (I) and h is any number, then kx^, ky^, kz^, kw^ is also a solution. 224 ADVANCED ALGEBRA The proof of this theorem is evident on substituting kx^ etc., in equations (I) and observing that the number A; is a factor of each equation. Thus if a system of n linear homogeneous equar tions has any solution distinct from the zero solution it has an infinite number of solutions. EXERCISES Solve for all the variables : x + y = a, a; + 3 y = 19, y + z = b, y-]-Bz = S, M + « = d, w + 3d = 11, V + x = e. V + 3ic = 15. z-\-y + w = a, Bx + y + z = 20, ^z + w + x = b, -a; + 4y + 3w; = 30, *w + a; + y = c, '6jc + 2 + 3iy = 40, x-{-y-\-z = d. 8y + 3z + 6w; = 50. 2/ + 2; + 5iy = ll, x — 2y + Sz — w = 6, ^ x + z-\- 4:W = llj ^y-2z + Sw-x = 0^ 'a: + y + 3M; = ll, ' z-2w-\-Sx-y = 0, x + z + Sy = 3S. w-2x + Sy - z = 6. 7. x + y-\-z + w = 2i, x + y + z + w = 60, a; + 2y + 3z-9w = 0, g x + 2y + 3z -\- ^w = 100, 3x-y -6z-\-w = 0, ' x-\-Sy -{■Qz-\-10w = 150, 2x + 3y--4«- 610 = 0. x + 42/ + lOz + 20mj = 210. CHAPTEE XIX PARTIAL FRACTIONS 202. Introduction. For various purposes it is convenient to f(x) express a rational algebraic expression ~j-\ ? § 11, as the sum of several fractions called partial fractions, which have the several factors of <^{x) as denominators and which have constants for numerators. If we write <j>(x) = {x — a)(x — P)--- (x — v), we seek a means of determining constants Aj B, ■, N such that for every value of x <l>{x) X — a X — p X — V ^ ^ If the degree of f(x) is equal to or greater than that of ^ (a;), we can write <i>(x) <}>(x) ^ ^ where Q (x) is the quotient and f^ (x) the remainder from dividing f(x) by <f>(x), and where the degree oi /^(x) is less than that of <^ (x). In what follows we shall assume that the degree of f(x) is less than that of <f> (x). In problems where this is not the case one should carry out the long division indicated by (2) and apply the principle developed in this chapter to the expression corre- sponding to the last term in (2). 203. Development when </>(x) = has no multiple roots. Let us consider the particular case f(x) ^ x + 1 <t> (x) (x -l)(x- 2) (x - 3)' We indicate the development required in form (1) of the last section, . . « ^ (x-l)(x-2)(x-3) x-1 x-2 x-S 225 226 ADVANCED ALGEBRA where A, B, and C represent constants which we are to deter- mine if possible. The question arises immediately, Are we at liberty to make this assumption? Are we not assuming the essence of what'we wish to prove, i.e. the form of the expansion? To this we may answer. We have written the expansion in form (1) tentatively. We have not proved it and are not certain of its validity. If, however, we are able to find numerical values of Ay B, and C which satisfy (1), we can then write down the actual development of the fraction in the form of an identity. If, on the other hand, we can show that no such numbers A,BjC satisfying (1) exist, then the development is not possible. Clear (1) of fractions, X + 1 = A(x - 2){x - ^)+ B{x -l){x - ^)-\- C (x -l)(x - 2) = {A + B-^C)x^-{6A +4.B-\-^C)x^QA+^B-\-2C. Since we seek values of A, B, and C for which (1) is identi- cally true for all values of x, equate coefficients of like powers of X in the last equation (Corollary II, p. 174). We obtain (2) (3) (4) (5) (2) A-\-B+ C = -5A _45-3C= 6A +SB-\-2C = = 0, -.1. Add (4) to (3) and we obtain Adding we obtain A -B- C= A+B-hC = 2A = A = = 2, :1 ^ From (^] 1 and (4) we obtain B=-3j C = :2. Tlma x + 1 1 XIlUo (x-l)(x -2)(a.-3) X-1 4- As a check we might clear of fractions and simplify. If equar tions (2), (3), and (4) had been incompatible, we should have con- cluded that we could not develop the fraction in form (1). PARTIAL FRACTIONS 227 We assume now for the general case ^(x) = (x - a)(x - P)--(x - v), and that the n roots a, )3, • • •, v are all distinct from each other. Let us consider the expression f(x)_ A B N <^{x) X — a X — p X — V ^ where A, B, •", N are constants. Let us assume for the moment the possibility of expressing ~y^ in terms of these partial frac- tions. We shall now attempt to determine actual values A, B,--, iV which satisfy such an identity. K we multiply both sides of the identity by <t>(x) = (x-a)(x-P)-'(x-v), we obtain f(x) = A(x-P)'"(x-v) + B(x-a)--'(x-v) + -" ^N(x-a)(x-P)"; f(x) is of degree not greater than n — 1, and consequently when written in the form of (1), p. 166, has not more than n terms. If we multiply out the righ1>hand member and collect powers of ar, we have an expression in a; of degree n — 1. By Corollary II, p. 174, this equation will be an identity if we can determine values ot Aj Bf "-,N which make the coefficients of x cm both sides of the equation equal to each other. Hence we equate coefficients of like jx>wers of x and obtain n equations linear in A, B, '",N which we can treat as variables. These equations have in general one and only one solution which we can easily deter- mine. The values of Aj B, • • • , iV obtained by solving these equa- tions we can substitute for the numerators of the partial fractions in (6). After making this substitution we can actually clear of • fractions the right-hand member of (6) and check our work by showing its identity with the left-hand member. There is no general criterion that we have applied to (6) to determine whether the n linear equations obtained lay equating coefficients of x have any solution or not. Hence in this general 228 ADVANCED ALGEBRA discussion it should be distinctly understood that assumption (6) holds when and only when these equations are solvable. In any particular case we can find out immediately whether the equations are solvable by attempting to solve them. If the num- bers Aj By ' • ', N do not exist, the fact will appear by our inability to solve the linear equations. As a matter of fact, one and only one solution always exists under the assumption of this section. If in (6) we assume that several of the symbols A, B, ■'■, N stand for expres- sions linear in" x, as, for instance, ax + b, we should then have a larger number of variables to determine than there are equations. Under these circumstances there is an infinite number of solutions of the equations. Thus if we should seek to express =^-7— as the sum of partial fractions where the numerators are not con- stants but functions of z, we could get any number of such developments. We have the following EtTLE. Factor <\>{x) into linear factors, as (x — a)(x — P)"-{x — v). Write the expression </)(a?) X — a X — P x — v Multiply both sides of the expression by <i>{x)y equate coefficients of like powers of x, and solve the resulting linear equations for A,B,-^-,N. Replace A, B, •", N hy these values and check hy substituting for X some number distinct from a, ^, "•, v. EXERCISES Separate into partial fractions ; (x-l)(x-2)x /J.2 _ 2 A B f* Solution : Assume = 1 1 — . (1^ {x-l){x-2)x a;-lx-2a; ^' Multiply by (x - 1) (x - 2) x, x2 - 2 = ^(x - 2)x + J5(x - l)x + C(x - l)(x - 2), 3fi-2 = {A+B-^C)x^-{2A+B + SC)z -f 2^. PARTIAL FKACTIOKS 229 Equating coefficients of like powers of x, A + B + C = l, 2A+B + SC = 0, 2 O = - 2. Hence C = — 1. Solving we obtain Thus A + Bz = 2, 2A + B = = 3, A = = 1 J5 = = 1 ■ ' +- 1 x^-2 1.1 1 (a;-l)(x-2)a; z-1 x-2 x Check : Let Substituting in (1), — - = — ~ -\ x = -A, -1 -6 r -■2 *^.- 1_ 6~ 1 2 -b' 5+1=1. 6 -6 2. x-1 x^ -\-Sx-^2 A 1 3 a;2 _ 2 X - 8 (i 4a;2 {x^-i){x-l) R a;2_3a; + l 3. " + '' 1. 2a;2-5x-3 5x 6x2-5x-l 2x2-1 (x2 + 3a; + 2)(x- -1) a;2 4-4 9. (X - 1) (X - 2) (X - 3) (X - 2) (X + 2) (X - 1) 204. Development when ^ (a?) = O has imaginary roots. In the preceding section no mention has been made of any distinc- tion between real and imaginary values oi a, p, ■ -- , v. In fact the method given is valid whether they are real or imaginary. It is, however, desirable to obtain a development in which only real numbers appear. Let us assume the development iS=-^ + -^ + - + ^^ + -^' (1) ^ {x) X — a X — p X — fji X — V where let us suppose that /x and v are the only pair of conjugate imaginary roots of <^ (.r), m and n being conjugate complex numbers. 230 ADVANCED ALGEBRA Let fi = a-\-ib, v = a — ib. Then adding the corresponding terms of (1), we obtain m n _ X (m -{- n) — a (m + n) b(m — n) x — a — ib x — a + ib (x — a)^ -{- b^ (x — af-\-b'^ ^ ^ Since /a and y are the only imaginary roots of <f> (x) = 0, the last term of (2) is real, as is also the entire right-hand member (§ 152). Hence, letting the numerator x(7}i-\-n)— a(m-\-n)-\- ib (m — n) — Mx + N, we have the development /M^^4_ _B_ ■ Mx + N <^(x) x-a x- ft "^ (ic _ a)2 -|. J2 W By complete induction we can establish this form of numerator where there is any number of pairs of imaginary roots of <f>(x) = 0. ^e have proved the form (3) where there is one pair of imagi- nary roots. Assuming the form where there are k pairs, we can prove it similarly where there are A: + 1 pairs. Hence we have the Theorem. If <l>(x) is facto7^ahle into distinct linear and quad- ratic factors, but the quadratic factors are not further reducible fix) into real * factors, then is separable into partial fractions of the form ^^ ^ A B Mx + N + ;;+•••+ o ■ ■ > X — a X — P a? -\- ^x + V where x^ + fix -\- v is an irreducible quadratic factor of <t>(x). This theorem is of course true only under the condition that the linear equations obtained in the process of determining the constants are solvable. It turns out, however, that in this case as in § 203 the linear equations obtained always have one and only one solution provided that the roots are all distinct. * A real /actor is one whose coefficients are all real numbers. PARTIAL FRACTIONS 231 EXERCISES Separate into partial fractions : x2+ 1 (X - 1) (x2 + X + 1) Solution : Assi^me X2+1 A Bx + C ...^ (X - 1) {X2 + X + 1) X - 1 X2 + X + 1 Multiply by (x -^{af^ + a; + Ij, - ' x2 + 1 = ^ (x2 + x + 1) + (Bx + 0) (« - 1). Collecting like powers of x, 3fi + l = {A + B)x^-{- {A- B+ C)x + A-C. Equating coefficients of x, A + B = l, A-B-{-C = 0, A-C = l. Add (2) and (3) and solve with (1), 2A-B=1 A-{-B=:l Substituting in (1), Substituting in (3), SA = 2, A = l B = h > (1) (2) (3) Thus X2 + 1 x-1 (x-l)(x2 + x + l) 3(x-l) 3x2 + 3x + 3 Check: Let x = — 1. 2 2 -2 Substituting, Reducing, „ X2 + a + 1 2-1 3 6 3 2 3-3+3 1. x2 + l x3 + 4x x2 + 4 x8-2x2 + 3x- X 2 (X + 3) (2 x2 - X X2 + 1 -4) X* + X2 + 1 5x^-1 x4 + 6x2 + 8* 1 (X - l)(3x2 + x + 6) x3 + 3x2-2x-16 Hint. Factor by synthetic divi- sion (see p. 178) . 232 ADVANCED ALGEBRA 205. Development when ^ (x) = (x — a)~. In this case the method given in the previous sections fails, as the equations for determining the values of the numerators are incompatible. If ^®^®* f{x) ^ a,x-' + a,x-' +... + a^_^ <f>(x) (x-ay ' ^ ^ we can separate into partial fractions as follows. Let X — a = y, that is, x = y -\- a, and substitute in (1). We obtain after collecting powers of y _ = 1 — -^ f_ — _, y y y y where the ^'s are constants. Eeplacing yhyx — a,wQ have th^ following development : A, /(f) = ^0 I ^1 I ^'^ I ....). (x — a)" X — a (x — ay (x — ay (x — ay EXERCISES Separate into partial fractions : - x2 + 2 a; - 1 (X-3)8 Solution : Assume — -— = H — ■ -(- — -. (1) (X - 3)8 jc - 3 (X - 3)2 (X - 3)8 ^ ' Multiply by (x - 3)8, x2 4- 2 X - 1 = ^ (x2 - 6x + 9) + 5(x - 3) + C Collecting powers of x, = Ax'^ + {B-6A)x-\-9A-BB-\-C. Equating coefficients of x, ^ = 1, 5-6^ = 2, 9A-SB + C = -1. Solving, B = 8, C = U. „ x2 + 2x-l 1,8 14 Hence — ' = 1 \- (X - 3)8 X - 3 (X - 3)2 (X - 3)8 Check : Let x = 1. ^ ^ ,.,,.. ,,, 2 1 8 14 1 „ 14 Substituting in (1), __ = — - + --_=_- + 2-— , 1 1 PARTIAL FRACTIONS 233 2. ^^^.. 3. (x-2)3 (2a; + 1)2 (cc - 4)3 g x^ + x + 1 g a;- g - 2 a;^ + 3 x + 1 (2x-l)*" ' (ax + 6)2* * (3x-2)3 206, General case. When <f>(x)= has real, complex, and multiple roots, we may use all the previous methods simultane- ously. Hence for this case we assume the expansion f(^ (x — a)--- (Xx^+ fxx -\- v) ■ " (x — tY X — a Ax2-f fjiX -\- V X — T (£c — t)' EXERCISES Separate into partial functions : - g^ + 2x2 + i8x-18 ' (x-l)(x2 + x + l)(x-2)2' Solution : x* + 2x2 + 18x-18 A Bx-VG D . E (X - 1) (x2 + X + 1) ^ - 2)2 X - 1 x2 + X + 1 X - 2 (X - 2)2 = ^(x2 + X + 1) (x - 2)2 + {Bx + C) (X - 2)2 (X - 1) + i)(x-l)(x-2)(x2 + x + l) + JSJ(X-1)(X2 + X + 1) = {A-hB+D)x^-\-{-3A-6B + C-2D-\-E)x^ -{-{A + 8B-5C)x2 + (-4E + 8C'-D)x + (4 ^ - 4 C + 2 D - ^). Equating coefficients of like powers of x, A+ B + D =1, (1) -SA-6B-\- C-2D + E = 0, (2) A-^8B-5C =2, (3) - 4 B + 8 C - D =18, (4) 4 A -4C + 2D-^ = -18. (6) Adding (2) and (5), (1) and (4), we have, together with (3), ^_5E-3C' = -18, (6) A-3B-\-SC = 19, (7) A-\-SB-6C = 2. (8) 234 ADVANCED ALGEBRA Adding all three equations, we find 3^ = 3, or A = l. Substituting in (3) and (7) and solving, we find C = S, B = 2. Substituting in (1), we find D = - 2. Similarly, from (6), ^ = 6. x* + 2a;2 + 18x-18 1 , 2a; + 3 2 , 6 Thus — — - = + — 4- (x-l)(x2 + a; + l)(x-2)2 x-1 x^ + x + 1 x-2 (a; -2)2 Check: Let x = — 1. go J 2 2 6 Substituting, -^-^ = __ + ____ + _, ¥ = 2i-i = -V-. 2. -^!±^. 3. X{X- 1)3 . X3 -f 2 X2 + 1 _ 4. o. X(X-1)3 g 2x3 + x + 3 y (X2 + 1)2 8 ^~^ 9 ■ (x + l)2(x + 2)' 10. ^^^ + ^ . 11. (a:2-l)(x + 2) 12. ^^-^^ + ^ . 13. (x - 8) (X - 9) .. 2x2 -3x- 3 .- 14. 15. 16. "^^ — tl 17. (2x-3)(6x2-6x + l) (a;2_3a;4.2)(x-3) l-x* 5X + 12 x(x2 + 4) 43X-11 30(x2-l) X3 - X - 1 X4-16 x2 + 6x-8 x3-4x 2 (x2 + X + 3) (2 a • + 1) x3 4-2x2-3x + 1 (x + 3)(x2-4x + 5) 13 -5x CHAPTEE XX LOGARITHMS 207. Generalized powers. If h and c are integers, we can easily compute h''. When c is not an integer but a fraction we can com- pute the value of ¥ to any desired degree of accuracy. Thus if 6 = 2, c = I, we have 2^ = V2* = VS, which we can find to any number of decimal places. If, however, the exponent is an irrar tional number as V2, we have shown no method of computing the expression. Since, however, V2 was seen (p. ^5) to be the limit approached by the sequence of numbers 1, 1.4, 1.41, 1.414, ••, it turns out that 5^^ is the limit approached by the numbers The computation of such a number as S^"*^ would be somewhat laborious, but could be performed, since 5^-^^ = 5**^ = V5^*\ Thus it is a root of the equation x^^ = S^'*^ and could be found by Horner's method, p. 197. We see in this particular case that 5^^ is the limit approached by a sequence of numbers where the exponents are the successive approximations to V2 obtained by the process of extracting the square root. In a similar manner we could express the meaning of 5", where i is a positive integer and c is any irrational number. Assumption. We assume that the laws of operation which we have adopted for rational exponents hold when the exponents are irrational. Thus 6c.6d= 6c+d^ ^= 6e-d, {Tbd)c= {ho)d= bcd^ where c and d are any numbers, rational or irrational. 236 236 ADVANCED ALGEBRA 208. Logarithms. We have just seen that when h and c are given a number a exists such that ¥ = a. We now consider the case where a and h are given and c remains to be found. Let a = 8, ^ = 2. Then if 2*^ = 8, we see immediately that c = 3 satis- fies this equation. If a = 16, 5 = 2, then 2" = 16 and c = 4 is the solution. If a = 10, 6 = 2, consider the equation 2" = 10. If we let c = 3, we see that 2^ = 8. If we let c equal the next larger integer, 4, we see 2* = 16. If then any number c exists such that 2*^ = 10, it must evidently lie between 3 andj:. To prove the exist- ence of such a number is beyond the^^ope of this chapter, but we make the following Assumption. There always exists a real number x which satisfies the equation b^ = a, (1) where a and h are positive numbers, provided b ^ 1. Since any real number is expressible approximately in terms of a decimal fraction, this number x is so expressible. The power to which a given number called the base must be raised to equal a second number is called the logarithm of the second number. In (1) X is the logarithm of a for the base b. This is abbreviated into X = \ogf,a. (2) Thus since 28 = 8, 102 = 100, 3- 2 = J, 40 = 1, we have 3 = logaS, 2 = logiolOO, - 2 = logg j, = log4l. The number a in (1) and (2) is called the antilogarithm. EXERCISES 1. In the following name the base, the logarithm, and the antilogarithm, and write in form (2). (a) 36 = 729. Solution : 3 = base, 6 = logarithm, 729 = antilogarithm, logs 729 = 6. (b) 2* = 16. (c) 38 = 27. LOGARITHMS 237 2. Find the logarithms of the following numbers for the base 3 : 81, 243, 1, i, ij. 3. For base 2 find logarithms of 8, 128, |, ^V 4. What must the base be when the following equations are true ? (a) log 49 = 2. (b) log81 = 4. (c) log 225 = 2 . (d) log 625 = 4. 209. Operations on logarithms. By means of the law expressed in the Assumption, § 207, we arrive at principles that have made the use of logarithms the most helpful aid in computations that is known. Theorem I. The logarithm of the prodicct of two numbers is the sum of their logarithms. Let log^a. = a;, logfeC = y. Then by (1) and (2), p. 236, h^ = a, ¥ = c. Multiply (by Assumption, § 207), or by (1) and (2), \og^a-c = x -\- y. Theorem IL The logarithm of the nth power of a number is n times the logarithm of the number. Let logi,a = Xy or b' = a. Raise both sides to the nth. power, (b^f. = b'^ = a% or logft a" = nx. Example. Iogiol00 = 2, logio 1000 = 3. By Theorem I, logio 100,000 = 5, which is evidently true, since 10^ = 100,000. 238 ADVANCED ALGEBRA Theorem III. The logarithm of the quotient of two is the difference between the logarithms of the numbers. Let logj, a = Xj log6C = 2/. Then b'' = a, b^ = c. Dividing, ^,«-y = -, c ^^ , a or log;, -==x — y. G Theorem IV. The logarithm of the real nth root of a nur.t- ber is the logarithm of the number divided by n. Let logj, a = Xj or b"" = a. 1 X Extract the nth root, (b'^y = ^^ = -Va, or logj -Va = - • n EXERCISES Given logio2 = .301, logio5 = .699, logio? = .8451, find 1. log(-s/f^- V5).* Solution : By Theorem I, log ( v^T^ • VS) = log v^ + log V5. By Theorems III and IV, = f log 7 + ^ log 5. Now f log 7 = f (. 8451) = . 50706, and i log 5 = |_(.699) = .349^ Adding, | log 7 + | log 5 = log ( Vt^ . VB) = .85656 2. log 40. 3. log 28. Hint. Let 40= 8-5= 28.6. 4. log 140. 5. log V280. 6. log V'35. 7. log ( V8 . v^ . V^). 8. log ( V5 . 78). 9. log ("t/Ie . Vli . 4^700), • Where no base is written it is assumed that the base 10 is employed. LOGARITHMS 239 210. Common system of logarithms. For purposes of compu- tation 10 is taken as a base, and unless some other base is indi- cated we shall assume that such is the case for the rest of this chapter. We may write as follows the equations which show the numbers of which integers are the logarithms. Since 10^ = 100,000 we have log 100,000 =A 10* = 10,000 log 10,000 = 4. 108 ^ IQQQ log 1000 = 3. 102 ^ IQQ log 100 = 2. 10^ = 10 log 10 = 1. 10« = 1 logl = 0. io-» = .i log.l = -1. 10-2 = .01 log .01 = -2. 10- « = .001 log .001 = -3. etc. etc, Assuming that as x becomes greater log x also becomes greater, we see that a number, for example, between 10 and 100 has a logarithm between 1 and 2. In fact the logarithm of any number not an exact power of 10 consists of a whole-number part and a decimal part. Thus since IQS < 3421 < lO^, log 3421 = 3. + a decimal. Since 10- 3 < .0023 < 10-2, log .0023 = - 3. + a decimal. The whole-number part of the logarithm of a number is called the characteristic of the logarithm. The decimal part of the logarithm of a number is called the mantissa of the logarithm. The characteristic of the logarithm of any number may be seen from the above table, from which the following rules are imme- diately deduced. 'f The characteristic of the logarithm of a number greater than unity is one less than the number of digits to the left of its decimal point. Thus the characteristic of the logarithm of 471 is 2, since 471 is between 100 and 1000; of 27.93 is 1, since this number is between 10 and 100; of 8964.2 is 3, since this number is between 1000 and 10,000. 240 ADVANCED ALGEBRA The characteristic of the logarithm of a number less than 1 is one greater negatively than the nwmher of zeros ^preceding the first significant figure. Thus the characteristic of the logarithm of .04 is — 2 ; of .006791 is — 3 ; of .4791 is - 1. It must constantly be kept in mind that the logarithm of a number less than 1 consists of a negative integer as a character- istic plus a positive mantissa. To avoid complication it is desir- able always to add 10 to and subtract 10 from a logarithm when the characteristic is negative. Thus, for instance, instead of writ- ing the logarithm - 3 + .4672 we write 10 - 3 + -4672 - 10, or 7.4672 — 10. This is convenient when for example we wish to divide a logarithm by 2, as by Theorem IV, § 209, we shall wish to do when we extract a square root. Since in the logarithm — 3 -f .4672 the mantissa is positive, it would not be correct to divide — 3.4672 by 2, as we should confuse the positive and negative parts. This confusion is avoided if we use the form 7.4672 - 10, and the result of division by 2 is 3.7336 - 5, or 8.7336 — 10. The actual logarithm which is the result of this division is — 2 + .7336. Theorem. Numhers with the same significant figures which differ only in the position of their decimal points, have the same mantissa. Consider for example the numbers 24.31 and 2431. Let 10^ = 24.31. . Then a = log 24.31. If we multiply both numbers of this equation by 100, we have 10^10^ = 10^+2^2431, or x-\-2 = log 2431. Thus the logarithm of one number differs from that of the other merely in the characteristic. In general numbers with the same sig- nificant figures are identical except for multiples of 10. Hence their logarithms differ only by integers, leaving their mantissas the same. Thus if log47120. = 4.6732, log47.12 = 1.6732, and log .004712 = - 3.6732, or 7.6732 - 10. LOGARITHMS 241 EXERCISES By §209, log V600 = 1.38905 2. log .06. 3. log (210)3. 4. log 5. log(4.2)4. 6. log "^2.1 7. log a , 567 Q , 13.23 9. log ^ 1.28 If log 2 = .3010, log 3 = .4771, log7 = .8451, find 1. logVOOO. Solution : log V600 = log V20 • 30 = ^ log 20 + i log 30. By the preceding theorem, log 20 = 1.3010, log 30 = 1.4771. i log 20= .6505 I log 30= .73855 (70)3 324 211. Use of tables. A table of logarithms contains the man- tissas of the logarithms of all numbers of a certain number of significant figures. The table found later in this chapter gives immediately the mantissas for all numbers of three significant figures. In the next section a method is given for finding the mantissa for a number of four figures.* Hence the table is called a four-place table. Before every mantissa in the table a decimal point is assumed to stand, but in order to save space it is not written. To find the logarithm of a number of three or fewer significant figures we apply the following Rule. Determine the characteristic hy rules in § 210. Find in column N the first two significant figures of the num- ber. The mantissa required is in the row with these figures. Find at the top of the page the last figure of the number. The mantissa required is in the column with this figure. When the first significant figure is 1 we may find the loga- rithm of any number of four figures by this rule from the table on pp. 248, 249 if we find the first three instead of the first two figures in column N. Thus the log 516. = 2.7126, log .00281 = - 3.4487, log 7400. = 3.8692, log 600. = 2.7782, log 50. = 1.6990, log 4.00 = .6021. 242 ADVANCED ALGEBRA EXERCISES Find the logarithms of the following : 1. 3. 2. 303. 3. .024. 4. 347. 6. .0333. 6. 1.011. 7. .202. 8. .0029. 9. .0001. 10. .00299. 11. 68400. 12. .0201. 212. Interpolation. We find by the preceding rule that, log 2440 = 3.3874, while log 2450 = 3.3892. If we seek the loga- rithm of a number between 2440 and 2450, say that of 2445, it would clearly be between 3.3874 and 3.3892. Since 2445 is just halfway between 2440 and 2450, we assume that its logarithm is halfway between the two logarithms. To find log 2445, then, we look up log 2440 and log 2450, take half (or .5) their difference, and add this to the log 2440. This gives log 2445 = 3.3874 + .5 x .0018 = 3.3883. If we had to find log 2442 we should take not half the difference but two tenths of the difference between the logarithms of 2440 and 2450, since 2442 is not halfway between them but two tenths of the way. This method is perfectly general, and we may always find the logarithm of a number of more than three figures by the following Rule. Annex to the proper characteristic the mantissa of the first three significant figures. Multiply the difference between this mantissa and the next larger mantissa in the table (called the tabular difference and denoted by D) by the remaining figures of the number preceded by a decimal point. Add this product to the extreme right of the logarithm of the first three figures, rejecting all decimal places beyond the fourth. In this process of interpolation we have assumed and used the principle that the increase of the logarithm is proportional to the increase of the number. This principle is not strictly true, though for numbers whose first significant figure is greater than 1 the error is so small as not to appear in the fourth place of the logarithm. For numbers whose first significant figure is less than 2 this error would often appear if we found the fourth place by interpolation. For this reason the table on pp. 248, 249 gives the logarithms of all such numbers exact to four figures, and in this part of the table we do not need to interpolate at all LOGARITHMS 243 EXERCISES Find the logarithms of the following : 1. 63.924. Solution: log 63.9 = 1.8055 2 Tabular difference = 4. 62230. 7. 20060. 10. 9.999. 13. 5.7828. 16. 3.1416. log 275 =2.4393 D= 16 6 2.4399 .4 6.4 log63.924 = 1.8057 iL 1.68 We add 2 to 1.8055 rather than 1, because 1.68 is nearer 2 than 1. In general we take the nearest integer. 2. 269.4. 3. 1001. 5. 392.8. 6. 9.365. 8. .4283. 9. .3101. 11. 82.93. 12. .05273. 14. .003011. 15. .002156. 17. 276.4 X 1.463. Solution : log 275.4 = 2.4399 log 1.463 = 0^652 By Theorem II, § 209, log (275.4 x 1.463) = 2.6051 18. 374.3 X 1396. 19. 1.46 x 237.2. 20. 469.1 X 63.92. 21. 47320. x .8994. 22. :5?!?1. ^8-^^ log .0372 =8.5705 -10 /)= 12 Solution: log. 03724 = 8.5710 - 10 5 _A log38.46 =L585^_ ^.^^J^'^ ,M 6.9860 - 10 7 .6 1.5850 7^ 23 5:^. 24 1^51??. .2364 ' 5.128 213. Antilogarithms. We can now find the product or quotient of two numbers if we are able to find the number that corresponds to a given logarithm. For this process we have the following KuLE. If the mantissa is found exactly in the table, the first two figures of the corresponding number are found in the column N of the same row, while the third figure of the number is found at the top of the column in which the mantissa is found. Place the decimal point so that the rules in § 210 are fulfilled. 244 ADVANCED ALGEBRA EXERCISES Find the antilogarithms of the following: 1. 3.7419. Solution: We find the mantissa .7419 in the row which has 55 in coT umn N. The column in which .7419 is found has 2 at the top. Thus the significant figures of the antilogarithm are 552. Since the characteristic is 3, we must by the rule in § 210 have four figures to the left of the point. Thus the number sought is 5520. 2. 1.3874. 3. 2.7050. 4. .6785. 5. 2.8414.* 6. 5.8831. 7. 1.5752. 8. 9.9112 - 10. ■ 9. 3.7251. 10. 6.3997. If the mantissa of the given logarithm is between two man- tissas in the table, we may find the antilogarithm by the following EuLE. Write the number of three figures corresponding to the lesser of the two mantissas between which is the given mantissa. Subtract this mantissa from the given mantissa, and divide this number by the tabular difference to one decimal place. Annex this figure to the three already found, and place the decimal point as the rules in § 210 require. It should be kept in mind that we may always add and subtract any integer to a logarithm. This is useful in two cases : First. When we wish to subtract a larger logarithm from a smaller ; Second. When we wish to divide a logarithm by an integer that is not exactly contained in the characteristic. Both these processes are illustrated in exercise 2 (1) following. EXERCISES 1. Find the antilogarithms of the following : (a) 2.3469. Solution : The mantissa 3469 is between 3464 and 3483. Hence D = 19. The mantissa 3464 corresponds to 222. To find the fourth significant figure of the antilogarithm, divide 3469 - 3464 = 5 by D = 19. Since 5 -^ 19 = .26, we annex 3 to 222. Hence the antilogarithm = 222.3. * We write - 2 + .8414 in the form 2.8414 to save space and at the same time to recall the fact that the mantissa is positive. LOGARITHMS 245 (b) 4.3147. (c) 1.5271. (d) 1.4216. (e) 1.6423. (f) 2.8791. (g) .7214. 2. Perform the following operations by logarithms. 1375 X .06423 (a) 76420 Solution : log 1375= 3.1383 log. 06423= 8.8077- 10 Adding (Theorem I, § 209), 11.9460 - 10 log 76420= 4.8832 Subtracting (Theorem III, § 209), log result = 7.0628 - 10 result = .001156. (b) (11)8. (c) iUUY- (d) 5871 -, - 9308. (e) (H)«. (f) (3f I)*-". (g) 7066 -. -5401. (h) 8308 X .0003769. (i) 3410 X .008763. 8.371 X 834.6 ,, , 37.42 X 11.21 ^■"^ 7309 ^^^ BBAl ■ ^1) 8/87xV7194 \ 98080000 Solution : log 87= 1.9395 By Theorem IV, § 209, i log 7194 = 1.9285 Adding, = 13.8680 - 10 log 98080000= 7.9916 By Theorem III, § 209, 3)25.8764 - 30 log result = 8.6255 - 10 result = .04222. Since in the subtraction in this problem we have to subtract 7 from 3, we add and subtract 10 to the minuend to avoid a negative logarithm. Since in the division by 3 we would have a remainder in dividing — 10 by 3, we add and subtract 20 so that 3 may be exactly contained in 30, the negative part of the logarithm. (m) ^. (n) V:06. (0) ^(.043)8. (P)^^. (q) (.21)§. (r) m V'lOO. (s) •^:o3. (t) ^100. (u) V(l. 563)3. (v) V.00614. (w) VHi- (X) ^0.9 VI!- (y) ■>J^-v^ (z) >y.47 VM- 214. Cologarithms. The logarithm of the reciprocal of a num- ber is called its cologarithm. When a computation is to be made 246 ADVANCED ALGEBRA in which several numbers occur in the denominator of a fraction, the subtraction of logarithms is conveniently avoided by the use of cologarithms. By our definition we have colog 25 = log 3jV = log 1 - log 25, Theorem III, § 209 log 1 = 10. - 10 log 25 = 1.3979 colog 25= 8.6021-10 Thus in dividing a number by 25 we may subtract the logarithm of 25, or what amounts to the same thing, add the logarithm of ^ij, which is by definition the cologarithm of 26. EuLE. The cologarithm of any number is found by subtract- ing its logarithm from 10 — 10. In the process of division subtracting the logarithm of a num- ber and adding its cologarithm are equivalent operations. EXERCISES Compute, using cologarithms. - 8 X 62.73 X .052 66 X 8.793 Solution : 2- MlVlf- 4. V38.462 - 15.382. g 5086 (.0008769)8 9802 (.001984)* * </: ' 'XVsox log 8= .9031 log 62. 73= 1.7975 log. 052= 8.7160-10 colog 56= 8.2518-10 colog 8.793 = 9.0559 - 10 log result = 27.7243 -30 result = .00^99 3. V1572 - 872. Hint . 1572 - 872 = (157 + 87) (157 - 87) = (244) (70). 6. V(27:5)2 - (3.483)2. 693 X .04692 03841 X (569.8)2 3 X 421.6 X .046.(200.1)^ (68.96)^ X 86.61 .09263 V^ 1416 X (5.638)2 (75)i LOGARITHMS 247 215. Change of base. We have seen that the logarithm of a number for the base 10 may be found to four decimal places in our tables. It is occasionally necessary to find the logarithm of a number for a base different from 10. For the sake of generality, we assume that the logarithms of all numbers for a base b are computed. We seek a means of finding the logarithm of any number, as x, for the base c ; that is, we seek to express log^a? in terms of logarithms for the base h. Suppose logc X = z, that is, c^ = x. Take the logarithm of this equation for the base h, and we have logftC^ = Z log^C = logftOJ. Then ^ log,a^ l0g6« If we let M = l0gf,Gj we have .. ^og,x (1) This number M does not depend on the particular number x, but only on the two bases. From (1) we see that we can find the logarithm of any number for the base c by dividing its logarithm for the base b by M. The number M is called the modulus of the new system with respect to the original one. EuLE. To find the logarithm of a member for a new base c, divide the common logarithm by the modulus of the system whose base is c. EXERCISES Find: 1. logs 21. solution: log321.>^-^; = lf^^f = 2.771. logioo .4771 2. log5 6. 3. logalS. 4. l0gi6 2. 5. logs 167. 6. logi8 237. 7. log2.i6l.41. 248 ADVANCED ALGEBRA N. 1 2 3 4 5 6 7 8 9 100 0000 0004 0009 0013 0017 0022 0026 0030 0035 0039 101 102 103 104 105 106 107 108 109 0043 0086 0128 0170 0212 0253 0294 0334 0374 0048 0090 0133 0175 0216 0257 0298 0338 0378 0052 0095 0137 0179 0220 0261 0302 0342 0382 0056 0099 0141 0183 0224 0265 0306 0346 0386 0060 0103 0145 0187 0228 0269 0310 0350 0390 0065 0107. 0149 0191 0233 0273 0314 0354 0394 0069 0111 0154 0195 0237 0278 0318 0358 0398 0073 0116 0158 0199 0241 0282 0322 0362 0402 0077 0120 0162 0204 0245 0286 0326 0366 0406 0082 0124 0166 0208 0249 0290 0330 0370 0410 110 0414 0418 0422 0426 0430 0434 0438 0441 0445 0449 111 112 113 114 115 116 117 118 119 0453 0492 0531 0569 0607 0645 0682 0719 0755 0457 0496 0535 0573 0611 0648 0686 0722 0759 0461 0500 0538 0577 0615 0652 0689 0726 0763 0465 0504 0542 0580 0618 0656 0693 0730 0766 0469 0508 0546 0584 0622 0660 0697 0734 0770 0473 0512 0550 0588 0626 0663 0700 0737 0774 0477 0515 0554 0592 0630 0667 0704 0741 0777 0481 0519 0558 0596 0633 0671 0708 0745 0781 0484 0523 0561 0599 0637 0674 0711 0748 0785 0488 0527 0565 0603 0641 0678 0715 0752 0788 120 0792 0795 0799 0803 0806 0810 0813 0817 0821 0824 121 122 123 124 125 126 127 128 129 0828 0864 0899 0934 0969 1004 1038 1072 1106 0831 0867 0903 0938 0973 1007 1041 1075 1109 0835 0871 0906 0941 0976 1011 1045 1079 1113 0839 0874 0910 0945 0980 1014 1048 1082 1116 0842 0878 0913 0948 0983 1017 1052 1086 1119 0846 0881 0917 0952 0986 1021 1055 1089 1123 0849 0885 0920 0955 0990 1024 1059 1093 1126 0853 0888 0924 0959 0993 1028 1062 1096 1129 0856 0892 0927 0962 0997 1031 1065 1099 1133 0860 0896 0931 0966 1000 1035 1069 1103 1136 130 1139 1143 1146 1149 1153 1156 1159 1163 1166 1169 131 132 133 134 135 136 137 138 139 1173 1206 1239 1271 1303 1335 1367 1399 1430 1176 1209 1242 1274 1307 1339 1370 1402 1433 1179 1212 1245 1278 1310 1342 1374 1405 1436 1183 1216 1248 1281 1313 1345 1377 1408 1440 1186 1219 1252 1284 1316 1348 1380 1411 1443 1189 1222 1255 1287 1319 1351 1383 1414 1446 1193 1225 1258 1290 1323 1355 1386 1418 1449 1196 1229 1261 1294 1326 1358 1389 1421 1452 1199 1232 1265 1297 1329 1361 1392 1424 1455 1202 1235 1268 1300 1332 1364 1396 1427 1458 140 1461 1464 1467 1471 1474 1477 1480 1483 1486 1489 141 142 143 144 145 146 147 148 149 1492 1523 1553 1584 1614 1644 1673 1703 1732 1495 1526 1556 1587 1617 1647 1676 1706 1735 1498 1529 1559 1590 1620 1649 . 1679 1708 1738 1501 1532 1562 1593 1623 1652 1682 1711 1741 1504 1535 1565 1596 1626 1655 1685 1714 1744 1508 1538 1569 1599 1629 1658 1688 1717 1746 1511 1541 1572 1602 1632 1661 1691 1720 1749 1514 1544 1575 1605 1635 1664 1694 1723 1752 1517 1547 1578 1608 1638 1667 1697 1726 1755 1520 1550 1581 1611 1641 1670 1700 1729 1758 150 1761 1764 1767 1770 1772 1775 1778 1781 1784 1787 N. 1 2 3 4 5 6 7 8 9 LOGARITHMS 249 N. 1 2 3 4 5 6 7 8 9 150 1761 1764 1767 1770 1798 1827 1855 1884 1912 1940 1967 1995 2022 1772 1775 1778 1781 1784 1787 151 152 153 154 155 156 157 158 159 1790 1818 1847 1875 1903 1931 1959 1987 2014 1793 1821 1850 1878 1906 1934 1962 1989 2017 1796 1824 1853 1881 1909 1937 1965 1992 2019 1801 1830 1858 1886 1915 1942 1970 1998 2025 1804 1833 1861 1889 1917 1945 1973 2000 2028 1807 1836 1864 1892 1920 1948 1976 2003 2030 1810 1838 1867 1895 1923 1951 1978 2006 2033 1813 1841 1870 1898 1926 1953 1981 2009 2036 1816 1844 1872 1901 1928 1956 1984 2011 2038 160 2041 2044 2047 2074 2101 2127 2154 2180 2206 2232 2258 2284 2049 2052 2055 2057 2060 2063 2066 161 162 163 164 165 166 167 168 160 2068 2095 2122 2148 2175 2201 2227 2253 2279 2071 2098 2125 2151 2177 2204 2230 2256 2281 2076 2103 2130 2156 2183 2209 2235 2261 2287 2079 2106 2133 2159 2185 2212 2238 2263 2289 2082 2109 2135 2162 2188 2214 2240 2266 2292 2084 2111 2138 2164 2191 2217 2243 2269 2294 2087 2114 2140 2167 2193 2219 2245 2271 2297 2090 2117 2143 2170 2196 2222 2248 2274 2299 2092 2119 2146 2172 2198 2225 2251 2276 2302 170 2304 2307 2310 2312 2315 2317 2343 2368 2393 2418 2443 2467 2492 2516 2541 2320 2322 2325 2327 171 172 173 174 175 176 177 178 179 2330 2355 2380 2405 2430 2455 2480 2504 2529 2333 2358 2383 2408 2433 2458 2482 2507 2551 2335 2360 2385 2410 2435 2460 2485 2509 2533 2338 2363 2388 2413 2438 2463 2487 2512 2536 2340 2365 2390 2415 2440 2465 2490 2514 2538 2345 2370 2395 2420 2445 2470 2494 2519 2543 2348 2373 2398 2423 2448 2472 2497 2521 2545 2350 2375 2400 2425 2450 2475 2499 2524 2548 2353 2378 2403 2428 2453 2477 2502 2526 2550 180 2553 2555 2558 2560 2562 2565 2567 2570 2572 2574 181 182 183 184 185 186 187 188 189 2577 2601 2625 2648 2672 2695 2718 2742 2765 2579 2603 2627 2651 2674 2697 2721 2744 2767 2582 2605 2629 2653 2676 2700 2723 2746 2769 2584 2608 2632 2655 2679 2702 2725 2749 2772 2586 2610 2634 2658 2681 2704 2728 2751 2774 2589 2613 2636 2660 2683 2707 2730 2753 2776 2591 2615 2639 2662 2686 2709 2732 2755 2778 2594 2617 2641 2665 2688 2711 2735 2758 2781 2596 2620 2643 2667 2690 2714 2737 2760 2783 2598 2622 2646 2669 2693 2716 2739 2762 2785 190 2788 2790 2792 2794 2817 2840 2862 2885 2907 2929 2951 2973 2995 2797 2799 2801 2804 2806 2808 191 192 193 194 195 196 197 198 199 2810 2833 2856 2878 2900 2923 294.: 2967 2989 2813 2835 2858 2880 2903 2925 2947 2969 2991 2815 2838 2860 2882 2905 2927 2949 2971 2993 2819 2842 2865 2887 2909 2931 2953 2975 2997 2822 2844 2867 2889 2911 2934 2956 2978 2999 2824 2847 2869 2891 2914 2936 2958 2980 3002 2826 2849 2871 2894 2916 2938 2960 2982 3004 2828 2851 2874 2896 2918 2940 2962 2984 3006 2831 2853 2876 2898 2920 2942 2964 2986 3008 200 3010 3012 3015 3017 3019 3021 3023 3025 3028 3030 N. 1 2 3 4 5 6 7 8 9 250 ADVANCED ALGEBRA N. 1 2 3 4 5 6 7 8 M 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 320r 21 22 23 24 25 26 27 28 29 3222 3424 3617 3802 3979 4150 4314 4472 4624 3243 3444 3636 3820 3997 4166 4330 4487 4639 3263 3464 3655 3838 4014 4183 4346 4502 4654 3284 3483 3674 3856 4031 4200 4362 4518 4669 3304 3502 3692 3874 4048 4216 4378 4533 4683 3324 3522 3711 3892 40(35 4232 4393 4548 4698 3345 3541 3729 3909 4082 4249 4409 4564 4713 3365 3560 3747 3927 4099 4265 4425 4579 4728 3385 3579 3766 3945 4116 4281 4440 4594 4742 3404 3598 3784- 3962 4133 4298 4456 4609 4757 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 32 33 34 35 36 37 38 39 4914 5051 5185 5315 5441 5563 5682 5798 5911 4928 5065 5198 5328 5453 5575 5694 5809 5922 4942 5079 5211 5340 5465 5587 5705 5821 5933 4955 5092 5224 5353 5478 5599 6717 5832 5944 4969 5105 5237 5366 5490 5611 5729 5843 5955 4983 5119 5250 5378 5502 5623 5740 6855 5966 4997 5132 6263 5391 5514 5635 5752 5866 5977 6011 6146 5276 5403 5527 6647 6763 5877 6988 5024 5159 5289 5416 6639 6658 5775 5888 5999 5038 5172 5302 5428 5551 5670 6786 6899 6010 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 42 43 44 45 46 47 48 49 6128 6232 6335 6435 6532 6628 6721 6812 6902 6138 6243 6345 6444 6542 6637 6730 6821 6911 6149 6253 6355 6454 6551 6646 6739 6830 6920 6160 6263 6365 6464 6561 6656 6749 6839 6928 6170 6274 6375 6474 6571 6665 6758 6848 6937 6180 6284 6385 6484 6580 6675 6767 6857 6946 6191 6294 6395 6493 6590 6684 6776 6866 6955 6201 6304 6405 6503 6599 6693 6786 6875 6964 6212 6314 6415 6513 6609 6702 6794 6884 6972 6222 6325 6425 6522 6618 6712 6803 6893 6981 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 52 53 54 55 56 57 58 59 7076 7160 7243 7324 7404 7482 7559 7634 7709 7084 7168 7251 7332 7412 7490 7566 7642 7716 7093 7177 7259 7340 7419 7497 7574 7649 7723 7101 7185 7267 7348 7427 7505 7582 7657 7731 7110 7193 7275 7356 7435 7513 7589 7664 7738 7118 7202 7284 7364 7443 7620 7597 7672 7745 7126 7210 7292 7372 7451 7528 7604 7679 7752 7135 7218 7300 7380 7469 7636 7612 7686 7760 7143 7226 7308 7388 7466 7543 7619 7694 7767 7152 7235 7316 7396 7474 7551 7627 7701 7774 60 7782 7789 7796 7803 7810 7818 7826 7832 7839 7846 61 62 63 64 65 66 67 68 69 7853 7924 7993 8062 8129 8195 8261 8325 8388 7860 7931 8000 8069 8136 8202 8267 8331 8395 7868 7938 8007 8075 8142 8209 8274 8338 8401 7875 7945 8014 8082 8149 8215 8280 8344 8407 7882 7952 8021 8089 8156 8222 8287 8351 8414 7889 7959 8028 8096 8162 8228 8293 8367 8420 7896 7966 8036 8102 8169 8235 8299 8363 8426 7903 7973 8041 8109 8176 8241 8306 8370 8432 7910 7980 8048 8116 8182 8248 8312 8376 8439 7917 7987 8055 8122 8189 8254 8319 8382 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 N. 1 2 3 4 5 6 7 8 9 LOGARITHMS 251 N. 1 2 3 4 5 6 7 8 9 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 72 73 8513 8573 8633 8519 8579 8639 8525 8585 8645 8531 8591 8651 8537 8597 8657 8543 8603 8663 8549 8609 8669 8555 8615 8675 8561 8621 8681 8667 8627 8686 74 75 76 8692 8751 8808 8698 8756 8814 8704 8762 8820 8710 8768 8825 8716 8774 8831 8722 8779 8837 8727 8785 8842 8733 8791 8848 8739 8797 8854 8745 8802 8859 77 78 79 8865 8921 8976 8871 8927 8982 8876 8932 8987 8882 8938 8993 8887 8943 8998 8893 8949 9004 8899 8954 9009 8904 8960 9015 8910 8965 9020 8915 8971 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 82 83 9085 9138 9191 9090 9143 9196 9096 9149 9201 9101 9154 9206 9106 9159 9212 9112 9165 9217 9117 9170 9222 9122 9175 9227 9128 9180 9232 9133 9186 9238 84 85 86 9243 9294 9345 9248 9299 9350 9253 9304 9355 9258 9309 9360 9263 9315 9365 9269 9320 9370 9274 9325 9375 9279 9330 9380 9284 9335 9385 9289 9340 9390 87 88 89 9395 9445 9494 9400 9450 9499 9405 9455 9504 9410 9460 9509 9415 9465 9513 9420 9469 9518 9425 9474 9523 9430 9479 9528 9435 9484 9533 9440 9489 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 91 92 93 9590 9638 9685 9595 9643 9689 9600 9647 9694 9605 9652 9699 9609 9657 9703 9614 9661 9708 9619 9666 9713 9624 9671 9717 9628 9675 9722 9633 9680 9727 94 95 96 9731 9777 9823 9736 9782 9827 9741 9786 9832 9745 9791 9836 9750 9795 9841 9754 9800 9845 9759 9805 9850 9763 9809 9854 9768 9814 9859 9773 9818 9863 97 98 99 9868 9912 9956 9872 9917 9961 9877 9921 9965 9881 9926 9969 9886 9930 9974 9890 9934 9978 9894 9939 9983 9899 9943 9987 9903 9948 9991 9908 9952 9996 100 0000 0004 0009 0013 0017 0022 0026 0030 0035 0039 N. 1 2 3 4 5 6 7 8 9 216. Exponential equations. Equations in which the variable occurs only in the exponents may often be solved by the use of tables of logarithms if one keeps in mind the fact that log oc^ = ic log a. EXERCISES Solve the following : 1. 10^-1 = 4. Solution : Taking the logarithm of both sides of the equation, we have (x-1) log 10 = log 4, or since log 10 = 1, a; = log 4 + 1 = .G021 + 1 = 1.0021. 252 ADVANCED ALGEBRA 2. 4* -3^ = 8, 2* . 81/ = 9. Solution : Taking the logarithms of the equations, we have a;log4 + ylogS = log 8, a: log 2 + y log 8 = log 9, or jc21og2 4-ylog3 = 31og2, (1) a;log2 + y31og2 = 21og3. Eliminate x. x21og2+ 2/log3 = 31og2 a;21og2 + y61og2 = 41og3 2/(log3 - 61og2) = 31og2 - 41og3 _ 31og2-41og3 _ 3 X .3010-4 x .4771 ^~ Iog3-01og2 ~ .4771 - 6 X .3010 .9030 - 1.9084 _ - 1.0054 _ 1.0054 ~ .4771 - 1.8060 ~ - 1.3289 ~ 1.3289* Perform this division by logarithms. log 1.0054 = 10.0023 - 10 log 1.3289= .1235 logy= 9.8788-10 y= .7565. Substituting in (1), a; = 31og2 - .7565 log3 _ .9030 - .7565 x .4771 2 log 2 ~ .6020 Compute .7666 x .4771 by logarithms. log. 7565= 9.8788-10 log. 4771= 9.6786-10 . log result = 19. 5574 - 20 result = .3609. „ .9030 -.3609 .5421 Hence x — = .6020 .6020 log.5421 = 19.7341 -20 log. 6020= 9.7796 - 10 log;c= 9.9545- 10 X = .9005. 3. 6* = 2. 4. 4y = 3. 5. 7*+« = 5. 6. 32x + i_5. 7.4^-1 = 6^+1. 8. 22a:+3_ 6^-1 = 0. a^.6y = m, -^ a^.6!' = w, c* . d?/ = 71. ' x-\- y — n. LOGARITHMS 253 -- 2^. 2^ = 222, -2 a2^-3.a3y-2:=a8. ' x-y = A. 8x + 2y = n. - 3 3* • 4y = 15562, .. ^^2^^ . ^/^^F^i = a\ 4^ . 52/ = 128000. 4,^:,— ^ 3 ^l)3x + 5 - ^sjb^y + 1 = 610. 217. Compound interest. If |1500 is at the yearly interest of 3%, the total interest for a year is $1500 • (0.03) = $45. The total sum invested at the end of a year would be $1545. Let, in general, P represent a sum of money in dollars. Let r represent a yearly rate of interest. Then P • r represents the yearly interest on P, and P-f p.r = P(r + 1) represents the total investment, principal and interest, at the end of a year. Similarly, P (r + 1) r is the second year's interest, and P{r + l)r + P(r + 1) = P(r2 + r + r + 1) = ^(^ + 1)^ is the total investment at the end of two years. In general, A = P (r + 1)" (1) is the total accumulation at the end of tz years. If we know r, P, and n, we can by (1) find A. If we take the logarithm of both sides of the equation, we have log A = log P -\- n log (r + 1), log^-logP ^ (2) log(r + l) ^ ^ Hence if we know A, P, and r, we can find n. If the interest is computed semiannually, we have as interest v at the end of a half-year P • ^ ' while the entire sum would be P (- 4- 1 ) . Reasoning as above, we find that if the interest is com- puted semiannually, the accumulation at the end of n years is ^=pg + l)" (3) Similarly, n= l"g -^ ' l»g f . (4) logg + l) 254 ADVANCED ALGEBRA 3 If the interest is computed k times a year, we liave at the end of n years A-P(^ + 1) , (5) Hogg + l) EXERCISES In such exercises as the following, four-place tables are not sufficiently exact to obtain perfect accuracy. In general, the longer the term of years and the more frequent the compounding of interest, the greater the inaccuracy. 1. If $1600 is placed at 3|% interest computed semiannually for 13 years, to how much will it amount in that time ? Hence ion: By formula (3), A = 4 -) 2n P = 1600, r = .03|, n = 13 36 A = 1600 {-^ \26 1600 \400 -)^=-Q^- log 1600= 3.2041 26 log407 = 67.8496 71.0537 261og400 = 67.6546 log^= 3.3991 A = $2507. Iog407 = 2.e096 26 156576 52192 67.8496 log 400 =2.6021 26 156126 52042 67.6546 2. After how long will $600 at 6% computed annually amount to $1000 ? Solution : By formula (2) we have _ log J. — logP log(r + l) A = 1000, P = 600, r = .06. log 1000 - log 600 3 - 2.7782 .2218 _ ^,. n = -^ = =^ = 8./0 years. log 1.06 .0253 .0253 ^ .76 year = .76 • 12 = 9.12 months. .12 month = .12 • 30 = 3.6 days. Thus n = 8 years 9 months 3.6 days. LOGARITHMS 255 In the following exercises the interest is computed annually unless the contrary is stated. 3. To what will $3750 amount in 20 years if left at 6% interest ? 4. To what sum will $25,300 amount in 10 years if left at 5% interest computed semiannually? 5. To what does $1000 amount in 10 years if left at 6% interest computed (1) annually, (2) semiannually, (3) quarterly ? 6. A sum of money is left 22 years at 4% and amounts to $17,000. How much was originally put at interest ? 7. What sum of money left at 4|% for 30 years amounts to $30,000 ? 8. What sum of money left 10 years at 4i% amounts to the same sum as $8549 left 7 years at 5% ? 9. If a man left a certain sum 11 years at 4%, it would amount to $97 less than if he had left the same sum 9 years at 5%. What was the sum ? 10. Which yields more, a sum left 10 years at 4% or 4 years at 10%? What is the difference for $1000 ? 11. Two sums of money, $25,795 in all, are left 20 years at 4f%. The difference in the sums to which they amount is $14,660. What were the sums ? 12. At what per cent interest must $15,000 be left in order to amount to $60,000 in 32 years ? 13. At what per cent must $3333 be left so that in 24 years it will amount to $10,000 ? 14. Two sums of which the second is double the first but is left at 2% less interest amount in 36 1 years to equal sums. At what per cent interest was each left ? 15. In how many years will a sum double if left at 5% interest ? 16. In how many years will a sum double if left at 6% interest computed semiannually ? 17. In how many years will a sum amount to ten times itself if left at 4% interest ? 18. In how many years will $17,000 left at 4^% interest amount to the same as $7000 left at 5|% for 20 years ? 19. On July 1, 1850, the sum of $1000 was left at 4i% interest. When paid back it amounted to $2222. When did this occur ? 20. Prove formulas (1), (3), and (5) by complete induction. CHAPTEE XXI CONTINUED FRACTIONS 218. Definitions. A fraction in the form e-¥f where a, &,•••, ^, •• • are real numbers, is called a continued fraction. We shall consider only those continued fractions in which tlie numerators b, d, /, etc., are equal to unity and in which the letters represent integers, as for example : »! H . written a^-\ — — When the number of quotients ^2? ^s? ^4? • • • is finite the frac- tion is said to be terminating. When the fraction is not terminat- ing it is infinite. We shall see that the character of the numbers represented by terminating fractions differs widely from that of the numbers represented by infinite continued fractions. We shall find, in fact, that any root of a linear equation in one variable, i.e. any rational number, may be represented by a terminating continued fraction, and conversely; furthermore, that any real irrational root of a quadratic equation may be represented by the simplest type of infinite continued fractions, and conversely. 219. Terminating continued fractions. If we have a terminat- ing continued fraction, where ^i, a2j • • • are integers, it is evident that by reducing to its simplest form we obtain a rational num- ber. The converse is also true, as we can prove in the following 256 CONTINUED FRACTIONS 257 Theorem. Any rational number may he expressed as a ter- minating fraction. Let - represent a rational number. Divide a by J, and let a^ be the quotient and c (which must be less than h) the remainder. Then (§ 26) , -^a, + - = a, + j Divide h by c, letting ^g t)® the quotient and d (which must be less than c) the remainder. Then - = «! H b a^ -\- d c Continuing this process, the maximum limit of the remainders in the successive divisions becomes smaller as we go on, until finally the remainder is zero. Hence the fraction is terminating. It is noted that the successive quotients are the denominators in the continued fraction. EXERCISES 1. Convert the following into continued fractions (a) ^VV- Solution: 247|77[0 77J247|3 231 16J77[4 64 13J16[1 13 3J13[4 12 1J313 3 The continued fraction is ZL-1 1 1 1 1. 247 ~ 3 + 4 + 1 + 4 + 3* 258 ADVANCED ALGEBRA (b) if. (c) ^Vt. (d) m- (e) iM- (f) Iff. (g) /jVf- (h) HH- (i) Hf. (J) mh 2. Express the following continued fractions as rational fractious. ^''^2 + 3 ^"^l + r ^"^-a + l- (d)l I. (e)i 11. (f)l 1 1. ^ ' a; + x ^ M+2 + 3 ^ ' 3 + 4 + 5 ,.11111 ,^111111 (g) ----- . (h) ------ . ^^2+4+2+4+2 ^^1+2+3+1+2+3 220. Convergents. The value obtained by taking only the first n ~ 1 quotients in a continued fraction is called the nth. convergent of the fraction. Thus iu the fraction 1+1111 2+3+2+6 : 1 is the first convergent, 1 3 1 + - = - is the second convergent, ■a 2t l + -=l+- = — is the third convergent, etc. 3 When there is no whole number preceding the fractional part of the continued fraction the first convergent is zero. Thus in 111 2 + 3 + 5 \ is called the second convergent. In the continued fraction ,1111 ^2 + ^3 + ^4 + «5 H let -i, -1, -i, . .. represent the successive convergents expressed 9.\ ?2 ^z as rational fractions. Then for the first convergent we have ^ = — > or^i = ai, g'i = l. 9.\ CONTINUED FRACTIONS 259 For the second convergent we have , 1 a^a^ + 1 J92 , H , ^ «! H = = — > or ^2 = «^2<^l + 1 = «2i?i + 1, ^2 <^2 2'2 !Z2 = «2 = ^22'l- For the third convergent we have ^2 + 1^ a3«^2 + 1 «^3«2 +1 2'3 «3 or i?3 = % (^2% + 1) 4- ^1 = ^3^2 +i?l, ?Z3 = «3«2 + 1 = %'Z2 + (1\- This indicates that the form of the r^th convergent is 9'n a«2'n-l + 2'»-2 ^ This is in fact the case, as we proceed to show by complete induction. We have already established form (1) for n = 2 and n = ^. We assume it for n = m, and will show that its validity for n = m -\-l follows. The (m + l)th convergent differs from the mth only in the fact that a^ -\ appears in the continued fraction in place of a^. In (1) replace n by m, and a„ by a^ H J and we have Pr 1+1 __ \ *^m+l/ '■^^ Um + ) 2'm-l + 2'm-2 \ «'m+l/ _ (Q^m-H<^>n + ^)Pm^l + Q^„. + 1 J^m - 2 (am+;iam + l).?m-l + «m + l?m-2 _ Q^m4-lKi>m-l+i>m~2)+i>»»-l «^». + l(am2'm-l + 9'm-2)+ 2'm-l which is form (1). 260 ADVANCED ALGEBRA EXERCISES 1. Express the following as continued fractions, and find the convergents. (a) !?. Solution : By the method already explained, we find that 30 _1 1 1 1 1 1 il~l + 2+l + 2 + l + 2' Here ai = 0, a2 = 1, as = 2, 04 = 1, as = 2, aQ = 1, a? = 2. The first convergent is evidently 0, the second is 1, and the third is 1 2 mi- ir ^. ^ - P4 CliPS +P2 1-2+1 3 The fourth convergent is — = -^^ — — = = -. Qi a^qs + ga 1-3 + 1 4 mu ^x^u . • P5 ttsP* + P3 2-3 + 2 8 The fifth convergent is — = —^ — ^ = = — • ?5 a^qi + ^3 2-4 + 3 11 aePn + P4 1-8 + 3 11 The sixth convergent is ^^ = The seventh convergent is ^6 tteO's + 5'4 1-11 + 4 15 Pi OtPg+Ps 2-11 + 8 30 (b) 3¥2. (c) If. (d) t¥t- (e) tVt- (f) t?7- Qi a^q& + q^ 2-15 + 11 41 (f) \'-'- (g)TVj. (b)TW^. (i)?\V U) mV 2. Find the value of the following by finding the successive convergents. ,,11111 ,^,11111 (a) - - - - -• (b) - - - - -. ^^2+1+2+1+2 ^'3+2+3+2+3 (c^ 1 i i i i i (d^ 1 1 1 1 1 1 ^^^2 + 3+1 + 1 + 3 + 2* ^^3 + 3_^34.34.34.3' (e^ 1 1 ' 1 1 1 1 (f\ 1 1 1 1 1 1 ^^^6 + 3 + 1 + 1 + 3 + 6' ^^1 + 34.5+5 + 3+1' 1 1 1 1 1 1 ^^^ (X - 1) + X + (X + 1) * ^ ^ X + X + x' 221. Recurring continued fractions. We have seen that e very- terminating continued fraction represents a rational number, and conversely. We now discuss the character of the numbers repre- sented by the simplest infinite continued fractions. A recurring continued fraction is one in which from a certain point on, a group of denominators is repeated in the same order. CONTINUED FRACTIONS 261 ----- ^ ^^ 3 + 2 + 3 + 2 + 3+2 + *"' 111111 1+2+3+1+2+3+'" are recurring continued fractions if the denominators are assumed to repeat indefinitely as indicated. That a repeating continued fraction actually represents a num- ber we shall establish in § 223. Unless this fact is proven, one runs the risk of dealing with symbols which have no meaning. If for certain continued fractions the successive convergents increase without limit, or take on erratic values that approach no limit, it is important to discover the fact. All the fractions that we discuss actually represent numbers, as we shall see. We shall consider only continued fractions in which every denominator has a positive sign. Theorem. Every recurring continued fraction is the root of a quadratic equation. T w • . 111111 Let, tor instance, a; = - - - - - - •••. ' ' a -\- b -\- c -\- a -\- b -\- G -[- Evidently the part of the fraction after the first denominator c may be represented by cc, and we have thus virtually the termi- nating fraction X a -\- b -\- c -\- X The second convergent is — The third convergent is ab -\-l q^ The fourth convergent, or a?, gives us ^jP4^ CT4^3+-^2 ^ (c+-a;)5+-l q^ ci^qz 4- S'2 (c + x) (ab +- 1) 4- fit Simplifying, we get (ab + l)x^-{- lc(ab -{-l)-{- a - b'jx - be -1 = 0, 262 ADVANCED ALGEBRA which is a quadratic equation whose root is a?, the value of the continued fraction. Since this equation has a negative number for its constant term it has one positive and one negative root. The continued fraction must represent the positive root, since we assume that the letters la, h, c represent positive integers. The quadratic equation whose root is a recurring continued fraction with positive denomina- tors will always have one positive and one negative root. The equation will be quadratic, however, whatever the signs of the denominators may be. The proof may be extended to the case where there are any number of recurring denominators or any number of denominators before the recurrence sets in. Since every real irrational root of a quadratic equation is a surd, our result is equivalent to the statement that every recurring continued fraction may be ex jft*essed as a surd. EXERCISES Of what quadratic equations are the following roots ? Express the con- tinued fraction as a surd. ill 1 1 * 2 + 3 + 2 + 3 + ""' Solution : Let x = - - . 2 + 3 + a: Then x = l 1 ^ + ^ 2+- 6+2X+1 3 + « or 2 0:2 + 6 X - 3 = 0. Solving this equation, we get - 3 + VT6 - 3 - Vl5 X\ = or X2 = 2 2 Since x^ is negative, xi must be the surd that is represented by the con- tinued fraction. 1111 _3 + Vl6 Thus 2. 2+3+2+3+ 2 1111 3IIII 1 + 2 + 1 + 2 + "*' ■ 3 + 2 + 3 + 2 + 6.2 + 1 1 1 1 . 2+1+2+1+ ••• m.r.u. = 2 + l^\^. ••. then ^-^=l^\^-- and . = 2 + i^-;^(^. -2) CONTINUED FRACTIONS 263 ^111111 silil • 1 + 2 + 3 + 1+2 + 3 + "" •3+1 + 3 + 1 + *"' « o 1 1 1 7. 3 + - - - .... ^3+4+5+ 8.1 + 1 1 1 1 3+4+3+4+ 9. 1 + i 1 1 i i 1 .... 1+2+3+1+2+3+ 10. i 1 1 -1 1 i .... 2+1+2+2+1+2+ 222. Expression of a surd as a recurring continued fraction. This is the converse of the problem discussed in the last sec- tion, and shows that recurring continued fractions and quadratic equations are related in the same intimate way that terminat- ing fractions and rational numbers (i.e. the roots of linear equa- tions) are connected. We seek to express an irrational number, as, for instance, V2, as a continued fraction. This we may do as follows. Since 1 is the largest integer in V2 we may write V2 = l + (V2-l) = l+(-^^^^. Rationalizing the numerator, we have V2 = i + --i — V2+-1 Since 2 is the largest integer in V2 +- 1 we have V2 = l+ 7^= 7 = 1 + 2+-(V2-l) o I (V2-I) Rationalizing the numerator V2 — 1, we have ^"^viTI ^ + 2+(V2-i) 264 ADVANCED ALGEBRA By continuing this process we continually get tlie denomi- nator 2. Thus 11 This process consists of the successive application of two opera- tions, and affords the EuLE. Express the surd as the sum of two numbers the first of which is the largest integer that it contains. nationalize the numerator of the fraction whose numerator is the second of these numbers. Repeat these operations until a recurrence of denominators is observed. This process may be applied to any surd, and a continued frac- tion which is recurring will always be obtained. We shall con- tent ourselves with a statement of this fact without proof. If the surd is of the form a — V^, a continued fraction may be derived for + V^ and its sign changed. Since the real roots of any quadratic equation cc^ + 2 a^x + ag = are surds of the form a ± V^, where a and b are integers, it appears that the roots of any such equation may be expressed as recurring continued frac- tions. It can be shown that the real roots of the general quad- ratic equation a-o^^ + ajcc -f ctg = may also be so expressed. EXERCISES 1. Express the following surds as recurrent continued fractions, (a) 2 + Vs. Solution : 2+V3 = 3 + (V3-l) = 3+ (1) = 34^^ = 3+ 2' V3+ 1 V3+I = 3 + ^^^ — = 3 + J = 3 + V3 + I ^ V3+I _ . V3-I 2 .2 "^2 3 + 1^ 3-1 =3 + ^^^ = 3 + 1 1 2(V3 + 1) V3 + I 2-|-(V3-l) 1 conti:nued fkactions 265 But since Vs — 1 is the same number that we have in (1), this fraction repeats from this point on, and we have 2+V3 = 8 + l 1 I 1 .... 1+2+1+2+ (b) V5. (c) Vl7. (d) V65. (e) V47. (f) Vli. (g) V23. (h) V3i. (i) Vl9. (J) V62. (k) V79. (1) V98. (m) V88. (n) V22. (0) Vis. (p) V59. (q) VlOl. (r) 7 + VII. (s) 8- -V3 (t) 3-V23. 2. Express as a continued fraction the roots of the following equations, y (a) a;2 - 7 X - 3 = 0. / (b) x^ + 2 a; - 6 = 0. (c) x2 + 3x - 8 = 0. (d) x2 - 4x - 4 = 0. 223. Properties of conver gents. The law of formation of con- vergents given in § 220 is valid whether the continued fraction is terminating or infinite. We should expect that in the case of an infinite fraction the successive convergents would give us an increasingly close approximation to the value of the fraction. This is indeed the fact, as we shall see. Theorem. The difference between the nth and (n + l)st con- vergents is ^ Mn + l We prove this theorem by complete induction. Let the continued fraction be a, H — — «2 + ^3 + »4 + Then the first and second convergents are respectively Then £2-a=(^i^L±l)_„, = i. 2'2 S'l »2 «2 (1) 266 ADVANCED ALGEBRA Since ^i = 1? 2'2 = ^2? we have ^zl±i _^ = izL^yHl for n = 1. We assume that the theorem holds for n — m^ that is, Prn±l _ ^ -y^m^m + i+gmPm + l ^ {zlJ}!^. We must prove that it holds for n = m -{• 1, Now since - ^"^^^ — - ^'" + ^ ^ Pm + l^m + 2 YPm + 29^m+l^ 2'm + 2 S'm + l 2'm + l2'»i + 2 our theorem reduces to proving that the numerator -- Pm+lQm + 2 ■^Pm + 2^m + l=(- l)""^"- (2) In the left-hand member of (2) set «m + 2'7m+l + ^m = 2'm + 2, (1)> § 220 and «^m + 2i^m + l+i^m=I>m + 2- Then -- Pm+li^m+^^m + l H- 2'm)f (^m+2P„.+ l +i?».)?n. + l =^m+l^m-^Pm<Im+l = - (Pm'Ln+ I - Pm + l^m) Corollary I. The difference between the successive convergents of a continued fraction with positive denominators approaches zero as a limit. Since q^ = a^q„_-^ + q^-^, evidently q^ increases without limit when n is increased, since to obtain a^ we add together positive numbers neither one of which can vanish. Thus we can find a value of n large enough so that — j and 1 . ^« . hence ? will be smaller than any assigned number, which is another way of stating that as n increases approaches zero as a limit. ^'•^'* "*" ^ CONTINUED FliACTIONS 267 Corollary II. The even convergents decrease, while the odd convergents increase, as n increases. We must show that PTn + 2 Pm 9.m + 2 5'm is negative or positive according as ni is even or odd. Adding and subtracting ? we have ?/m + 1 Pm + 2 Pm (Pjn±2 Pin±\ i , m+l 2'm + 2 ^m Vlm + 2 <! /Pjn±l_ln\ X^m + l ^mj _ (-1)"»+^ (-1)"* + ^ By Corollary I, the denominator of the first fraction exceeds that of the second. Hence when m is odd the sum in the last member of the equation is positive, and when 7n is even the sum is negative. We now se* that any recurring fraction of the type considered in § 221 actually represents a number in the sense of § 74. We have seen that the successive odd convergents continually increase, while the even convergents continually decrease, until the differ- ence between a pair of them is very small. Such sequences of numbers we have seen (§73 ff.) define real numbers. 224. Limit of error. We are now in a position to state a maxi- mum value for the error made in taking any convergent of a con- tinued fraction for the fraction itself. Theorem. The maximum limit of error in taking the nth convergent for the continued fraction is less than Since by the theorem of the last section the value of the frac- tion is between any pair of consecutive convergents, it must differ from either of these convergents by less than they differ from each other, that is, by less than 268 ADVANCED ALGEBRA EXERCISES Find a convergent that differs by less than .001 from each of the following ; 1. V6. Solution : V6 = 2 + ( V6 - 2) = 2 + ^_~^ = 2 + V6 + 2 V6'+2 2 = 2 + i 6-4 =2 + 5 1 =^ + lj_l Since the last surd repeats the one in the first equation we have V6 = 2 + i 1 1 1 .... 2+4+2+4+ Since Pi _ 2 . J92 _ 5 . Pa _ 4 • 5 + 2 _ 22 . gi ~ 1 ' 52 ~ 2 ' ^3 4 . 2 + 1 ~ 9 ' P4 2-22 + 5 49. ps 4-49 + 22 g4~2.9 + 2~20' 95~4-20 + 9~ 218 89 1-^-1 <.001. 54^5 20 - 89 1780 we see by § 224 that |f satisfies the condition, of the problem. 2. V7. 3. V46. , 4. Vs. 5. Vl9. 6. V36. 7. V32. 8. V6l. 9. V65. 10. S+V2.- 11. V99. 12. Vil. 13. Vl3. 14. The number ir has the value 3. 14169. Find by the method of continued fractions a series of convergents the last of which differs from this value by less than .OCioi. CHAPTEE XXII INEQUALITIES 225. General theorems. We say that a is greater than h when a — 6 is positive. If a — h is negative, then a is less than h. Thus any positive number or zero is greater than any negative number. As we distinguished between identities and equations of condition in § 53, so in this discussion we observe that some statements of inequality are true for any real value of the letters, while others hold for particular values only. The former class may be called unconditional inequalities, the latter conditional. Thus a2 > — 1 is true for any real value of a and is unconditional, while a; — 1 > 2 only when a; is greater than 3 and is consequently conditional. The two inequalities <* > 5, g'> d are said to have the same sense. Similarly, a <b, c <d have the same sense. The inequal- ities a>h, c <id have a different sense. Theorem I. Any positive number may he added to, subtracted from, or midtiplied by both numbers of an inequality without affecting the sense of the inequality. Let a> b, that is, let a — b = k, where k is a positive number. If m is a positive number, evidently a ± m — (b ± m) = k, or a ±m> b±m. Similarly, ma — mb = mk, or ma > mb. The other statements of the theorem are proved similarly. Corollary. Terms may be transposed from one side of an inequality to the other as in the case of equations. 269 270 ADVANCED ALGEBRA Let a> b -{- G. Subtract c from both sides of the inequality and we obtain by Theorem I a — c> h. Theorem IL If the signs of both sides of an inequality are changed, the sense of the inequality must be reversed, that is, the > sign must be changed to <, or conversely. Let a> b, that is, let a — b = k, where ^ is a positive number. Then -a-\-b=-k, or (— «)—(—&) = — 7c, that is, by definition, — a < — b. EXERCISES Prove that the following identities are tnie for all real positive values of the letters. 1. a2 + 62 > 2 ah. Solution : (a — h)^ is always positive. Thus a2 - 2 a6 + 62 = a2 + 62 _ 2 a6 is positive. That is, a2 + 62 > 2 ah. 2. 3(a3 + 63)>a26 + a62. 3. a2 + 62 + c2 > a6 + ac + he. 4. (6 + c) (c + a) (a + 6) > 8 ahc. 5. (a + 6 + c) (pfi + 62 + c2) > 9a6c. 6. 62c2 + c2a2 + a262 > a6c (a + 6 + c). 7. 3(a8 4- 63 + c8) > (a + 6 + c) {ah + 6c + ca). 8. V(x + xi)2 + (y + 2/i)2 < Vx2 + 2/2 + y/x^^ + yi^. 9. If a2 + 62 = 1, x2 + y2 = 1^ prove that ax-\-hy< 1. 10. (a + 6 - c)2 + (a + c - 6)2 + (6 + c - a)2 > a6 + 6c 4- ca. 11. Show that the sum of any positive number (except 1) and its reciprocal is greater than 2. 12. Prove that the arithmetical mean of two unequal positive numbers always exceeds their geometrical mean. INEQUALITIES 271 226. Conditional linear inequalities. If we wish to find the values of x for which ax -\- b < c, (1) where a, h, and c are numbers and a is positive, we may find such values by carrying out a process similar to that of solving a linear equation in one variable. By the corollary, § 225, we have from (1) ax < c — b. By Theorem I, § 225, x <- 227. Conditional quadratic inequalities. We have already shown in § 116 that the quadratic expression ax^ -\- bx -\- c is posi- tive or negative, when the equation ax^ -{- bx -{- G = (1) has imaginary or equal roots, according as a is positive or nega- tive. If the equation has distinct real roots, the expression is positive or negative for values between those roots according as a is negative or positive. This we may express in tabular form as follows, for all values of x excepting the roots of (1), for which of course the expression vanishes. a 62 _ 4 ac dx^ -\- bx + c + -orO Always + - -orO Always — + + - for X between roots, + for other values - + + for X between roots, - for other values This enables us to answer immediately questions like the following : Example. For what values of cc is — 2 ic^+a; > — 3 ? By the corollary, § 225, this is equivalent to the question, For what value ofxis —2x^-^x-\-S>0? Here 62 — 4 ac = 1 -}- 24 = 25 is positive. The roots of the equation — 2 a;2 4- a; + 3 = are ic = — 1, a; = f . Thus by our table this expression is positive for all values of x between — 1 and |. 272 ADVANCED ALGEBRA EXERCISES For what values of x are the following inequalities valid ? 1. 2x-3>0. 3. _x-l>7. - 9x 4 ^ 3 7 7. .12x.+ .3<1.3. 9. 3<5x-2. 11. x2-8x + 22>6. 13. 2x2-3x>5. 15. 2x2 -4x< -2. 17. -3x2 + 2x<2. 19. 5x2-8x<l. 21. 3x2>3x-3. 2. 4x-7>l. 4. - 3 X + 8 < 3. 6. 3 ^4^5 8. 3-4x>2. 10. .s<i|-. 12. x2 + 3x-2>l. 14. -3x2-4x>8. 16. 3x2-9x>-6. 18. - x2 + 6 X > 9. 20. x2 < X - 1. 22. 3x>2x2-4. CHAPTEE XXIII VARIATION 228. General principles. The number x is said to vary directly as the number y when the ratio of ic to 2/ is constant. , This we symbolize by i» oc 2/, or ^ = A^, (1) where A; is a constant. Thus if a man walks at a uniform speed, the distance that he goes varies directly as the time. If the length of the altitude of a triangle is given, the area of the triangle varies directly as the base. The volume of a sphere varies directly as the cube of its radius. The number x is said to vary inversely as the number y when X varies directly as the reciprocal of y. Thus x varies inversely as y when 05 oc -5 OY - — xy =^ Uy (2) y i y where A: is a constant. Thus the speed of a horse might vary inversely as the weight of his load. The length of time to do a piece of work might vary inversely as the nmnber of laborers employed. The intensity of a light varies inversely as the square of the distance of the light from the point of observation. If I repre- sents the intensity of light and d the distance of the light from the point of observation, we have Zoc|. or| = ZcZ^ = A5, (3) where A; is a constant. - ' - 273 274 ADVANCED ALGEBRA The number x is said to vary jointly as y and z when it varies directly as the product of y and z. Thus x varies jointly as y and z when X QC yz, or yz h (4) where Aj is a constant. Thus a man's wages might vary jointly as the number of days and the number of hours per day that he worked. The number x is said to vary directly as y and inversely as z when it varies directly with -• Thus the force of the attraction of gravitation between two bodies varies directly as their masses and inversely as the squares of their distances. If m represents the masses of two bodies, d their distance, and G the force of their attraction due to gravity, then m Goc^, or = k. (P) EXERCISES 1. If a varies inversely as the square of 6, and if a = 2 when 6 = 3, what is the value of a when 6 is 18 ? Solution: By (3), a62 = k. We can determine k by substituting a = 2, 6 = 3. 2 . 9 = A;. 18 = A:. Then a . (18)2 = 18, or a = tV- 2. The volume of a sphere varies as the cube of its radius. A sphere of radius 1 has a volume 4.19. What is the volume of a sphere of radius 3 ? Solution: Let F represent the volume and r the radius of the sphere. Then by (1), \ = k. 4.19 Determine k by substituting. Then = k. A; = 4. 19. V V -- = — = 4.19. (3)8 27 F= 113.13. VARIATION 275 3. ltxyxx + y, and a; = 1 when y = 1, find x when y = 8. 4. The area of a circle varies as the square of the radius. If a circle of radius 1 has an area 3.14, find the area of a circle whose radius is 21. 5. Find the volume of a sphere whose radius is .2. Hint. See exercise 2. 6. The volume of a circular cylinder varies jointly with the altitude and the square of the radius of the base. A cylinder whose altitude and radius are each 1 has a volume of 3. 14. Find the volume of a cylinder whose altitude is 16 and whose radius is 3. 7. The weight of a body of a given material varies directly with its volume. If a sphere of radius 1 inch weighs f of a pound, how much would a ball of the same material weigh whose radius is 16 inches ? 8. The distance fallen by an object starting from rest varies as the square of the time of falling. If a body falls 16 feet in 1 second, how far will it fall in 6 seconds ? 9. A body falls from the top to the bottom of a cliff in 3| seconds. How high is the cliff ? 10. A triangle varies in area jointly as its base and altitude. The area of a triangle whose base and altitude are each 1 is ^. What is the area of a triangle whose base is 16 and altitude 7? 11. If 6 men do a piece of work in 10 days, how long will it take 5 men to do it? 12. If 3 men working 8 hours a day can finish a piece of work in 10 days, how many days will 8 men require if they work 9 hours a day ? 13. An object is 30 feet from a light. To what point must it be moved in order to receive (a) half as much light, (b) three times as much light ? 14. The weights of objects near the earth vary inversely as the squares of their distances from the center of the earth. The radius of the earth is about 4000 miles. If an object weighs 150 pounds on the surface of the earth, how much would it weigh 6000 miles distant from the center ? CHAPTEE XXIV PROBABILITY 229. Illustration. If a bag contains 3 wlaite balls and 4 blact balls, and 1 ball is taken out at random, what is the chance that the ball drawn will be white ? This question we may answer as follows : There are 7 balls in the bag and we are as likely to get one as another. Thus a ball may be drawn in 7 different ways. Of these 7 possible ways 3 will produce a white ball. Thus the chance that the ball drawn will be white is 3 to 7, or f. The chance that a black ball will be drawn is f . 230. General statement. It is plain that we may generalize this illustration as follows : If an event may happen in ^ ways and fail in q ways, each way being equally probable, the chance or probability that it will happen in one of the ^ ways is V The chance that it will fail is p + q (1) (2) The sum of the chances of the event's happening and failing is 1, as we observe by adding (1) and (2). The odds in favor of the event are the ratio of the chance of happening to the chance of failure. In this case the odds in favor are I. (3) q The odds against the event are -. P 276 PKOBABILITi 277 EXERCISES 1. K the chance of an event's happening is ^^^ what are the odds in its favor ? v 1 Solution : By (1), -^— = — . Hence lOjp = p + g, or 9p = g, «i 1 or - = - , which by (3) are the odds in favor. q 9 2. From a pack of 52 cards 3 are missing. What is the chance that they are all of a particular suit P Solution : The number of combinations of 62 cards taken 3 at a time is C52, 3 = — ' This represents p ■{■ q. The number of combinations of the 1 * 2 • 3 13 • 12 • 11 13 cards of any one suit taken 3 at a time is C13, 3 = ^ „ • This repre- sentep, 13 • 12 • 11 ^ p 1-2.3 13.12-11 11 11 Thus — =- — = = = ■ = . p-\-q 62 » 61 • 60 62.61-60 17-60 860 1-2-3 3. What is the chance of throwing one and only one 6 in a single throw of two dice ? Solution : There are 36 possible ways for the two dice to fall. This repre- sents p + g. Since a throw of two sixes is excluded there are 6 throws in which each die would be a 6, that is, 10 in all in which a 6 appears. This represents p. P 10 6 Thus — - — = — = — p + q 36 18 4. A bag contains 8 white and 12 black balls. What is the chance that a ball drawn shall be (a) white, (b) black ? 5. A bag contains 4 red, 8 black, and 12 white balls. What is the chance that a ball drawn shall be (a) red, (b) white, (c) not black ? 6. In the previous problem, if 3 balls are drawn, what is the chance that (a) all are black, (b) 2 red and 1 white ? 7. What is the chance of throwing neither a 3 nor a 4 in a single throw of one die ? 8. What is the chance in drawing a card from a pack that it be (a) an ace, (b) a diamond, (c) a face card ? 278 ADVANCED ALGEBRA 9. Three cards are missing from a pack. What is the chance that they are (a) of one color, (b) face cards, (c) aces ? 10. A coin is tossed twice. What is the chance that heads will fall once? 11. The chance that an event will happen is f. What are the odds in its favor ? 12. The odds against the occurrence of an event are |. What is the chance of its happening ? 13.' What is the chance of throwing 10 with a single throw of two dice ? 14. A squad of 10 men stand in line. What is the chance that A and B are next each other ? 15. What is the chance that in a game of whist a player has 6 trumps ? 16. What is the chance that in a game of whist a player holds 4 aces ? CHAPTER XXV SCALES OF NOTATION 231. General statement. The ordinary numbers with which we are acquainted are expressed by means of powers of 10. Thus 263 = 2 . 10^ + 6 . 10^ + 3. This is the common scale of notation, and 10 is called the radix of the scale. In a similar manner a number might be expressed in any scale with any radix other than 10. If we take 6 as the radix, we shall have as a 'number in this scale, for instance, 543 = 5 • 62 + 4 • 6 4- 3. In this scale we need only and five digits to express every positive integer. In general, if r is the radix of a scale of notation, any positive integer N will be denoted in this scale as follows : N=ao7^ + a^r--^ + a^r^-^ + . . . + o,^. (1) Theorem. Any positive . integer may he expressed in a scale of notation of radix r. Suppose we have a positive integer N. Let ?•" be the highest power of r that is contained in N. Then N = aoT- + iV'i, where N^ is less than r". Suppose that on dividing iVj by r"-^ we obtain ^^ ^ ^^^.„_i ^ ^^^ where N^ is less than r"~\ Then N = a^r^ + air«-i + N^. Proceeding in this manner we obtain finally N = aor"" -h 0^1^""^ H h a„, where the a's are positive integers less than r, or perhaps zeros. 279 280 ADVANCED ALGEBRA One observes that the symbol 10 indicates the radix in any system. In this general scale we need a and r — 1 digits to express every possible number. 232. Fundamental operations. In the four fundamental operar tions in the common scale we carry and borrow 10 in computing. In computing in a scale of radix 6, for instance, we should carry and borrow 6. If the radix were r, we should carry or borrow r. Thus let r = 6. Then 4 + 5=1-6+3=13. Similarly, 5-3=2. 6 + 3 = 23. This is precisely analogous to our computation in the common scale, where, for instance, we would have 9 + 8 = 1 • 10 + 7 = 17, or 6 • 7 = 4 - 10 + 2 = 42. EXERCISES Perform the following operations. 1. 2361 + 4253 + 2140 ; r = 7. 2361 4253 2140 - 12114 In this process, since 3 + 1=4 and is less than the radix, there is nothing to carry. The next column gives 6 + 5 + 4= 15 =2-7+1, hence we write down 1 and carry 2. The next column gives 3 + 2+1 + 2=8=1-7 + 1, hence we write down 1 and carry 1. Finally we get 2 + 4 + 2 + 1 = 9=1-7 + 2, hence we write 12. 2. 4602-3714; r = 8. 4602 3714 Since we cannot take 4 from 2 we horrow one from the next place. Since the radix is 8 this amounts to 8 units in the first place. We then subtract 4 from 8 + 2, which leaves 6. In borrowing 1 from that digit is really reduced to 7 and the preceding digit to 5 ; then subtracting 1 from 7 we get 6. Since we cannot take 7 from 5 we borrow 8 again and take 7 from 5 + 8= 13, which leaves 6. Since 1 has been borrowed from the 4 we see the subtraction is complete since 3-3=0. 3. 4321 . 432 ; r = 6. 4821 432 14142 24013 33334 4143222 In multiplying by 2 we have nothing to carry until we multiply 3 by 2. This gives 6=1-5 + 1. Hence we put down 1 and carry 1 to the product of 2 and 4. The addition of the partial products is carried out as in exercise 1. SCALES OF NOTATION 281 4. 32130 H- 43 ; r = 6. 43 1 32130 1 430 300 213 213 00 In making an estimate for the first figure in the quotient we divide 32 by 4, keeping in mind that for this purpose 32=3-64-2. Thus 4 is contained in 20 just 5 times, but since our entire divisor is 43 we take 4 as the first figure in the quotient. The multipli- cations are of course performed as in exercise 3, excepting that here 6 is the radix. 5. 4361 + 2635 + 5542 ; r = 1. 6. 5344 - 3456 ; r = 7. 7. 2340 . 4101 ; r=6. 8. 6435 • 35 ; r = 7. 9. 2003455 - 403 ; r = 6. 10. 344032 - 321 ; r = 5. 11. 534401 - 443524 ; r = 6. 12. 425 + 254 + 542 + 452 ; r=6. 233. Change of scale. If we have a number in the scale of radix r, we may find the expression for that number in the com- mon scale by writing the number in form (1), § 231, and carrying out the indicated operations.. Example. Convert 4635, where r = 7, into the ordinary scale. 4635 = 4- 73 + 6. 72 + 3-7+5 = 4 • 343 + 6 . 40 + 3 . 7 + 5 = 1692. If we have a number in the common scale, we may express it in the scale with radix r as follows : If the number is N, we have to determine the integers aQ, a^, • • , a„ in the expression N = ao7^ + «^i^""^ H h a^-iT + ««• (1) Divide (1) by r. We have :^= aor"-i + a,r-' H- a^,_, + - = iV' +-; r r r that is, the remainder a^ of this division is the last digit in the expression desired. Divide N' by r and we obtain — = JSJ-" = a^r^-^ + a^r^-^ H -|- ^^^^ ; r r 282 ADVANCED ALGEBRA that is, the remainder from this division is the next to the last digit in the desired expression. Proceeding in this way we obtain all the digits fl^„, c^„ _ i, • • • , ^i, «o- Example. Express 37496 in the scale with radix 7. 7) 37496 7 ) 5356 remainder 4 1 ) 765 remainder 1 • 7 ) 109 remainder 2 7 ) 15 remainder 4 7 )2 remainder 1 remainder 2 The number in scale r = 7 is 214214. To change a number from any scale rj to any other scale r^j we may first change the number to the scale of 10 and then by the process just given to the scale r^. The process indicated in the preceding example may be employed directly to change from any scale to any other, provided the division is carried out in the scale in which the number is given. One of these methods may be used to check the other. Example. Change 34503 from scale r = 6 to one in which r = 9. 34503 = 3. 6* + 4- 63 +5. 62 + 3 = 4935 in scale of 10. ♦ 9 )4935 9 ) 548 remainder 3 9 )60 remainder 8 9)6 remainder 6 remainder 6 Thus 34503 in scale of 6 becomes 6683 in scale of 9. Check: 9)34503 9)2312 remainder 3 ^" carrying out this division it must ■ X • A Q ^® kept in mind that the dividends are in 92140 remamder 8 g^^jg ^f g^ ^l^jle ^^Q remainders are to be 9) 10 remainder 6 in scale of 9. remainder 6 234. Fractions. In the ordinary notation we express fractional numbers by digits following the decimal point. This notation may also be used in a scale with any radix. SCALES OF NOTATION 283 Thus the expression .5421 stands for 10 102 ^ 10« 10* in the common scale. In the scale with radix r it stands for ry% ry%** n%9 niT The process of changing the scale for fractions is performed in accordance with the same principles as are employed in the change of scale for integers. The following examples suffice to illustrate it. Example 1. Express .5421 in the scale of 6 as a decimal fraction. .6421 = 5 + 1+1+1 6 62 68 64 5.63 + 4.62+2-6+1 1237 ^,,, 6* 1296 Example 2. Express .439 as a fraction for radix 6. Let .439 = ^ + A + l + l + .... 6 62 68 6* Multiplying by 6, 2.634 = a + ^ + ^ + ^ + • . . . b o-' 6' Thus a = 2 and we have .634 = - + — f- 1 . 6 ^ 62 68 c d Multiplying by 6, 3.804 = 6 + - + — + ... . 6 62 c d Thus 6 = 3 and we have .804 = - + — + ••. . 6 62^ Multiplying by 6, 4.824 = c + - + •• .. 6 d Thus c = 4 and we have .824 = - + .... 6 Multiplying by (T, 4.944 = d + • . .. Thus d = 4. The fraction in scale of radix 6 is then .2344 • • «„ 284 ADVANCED ALGEBRA EXERCISES 1. Express the following as decimal fractions, (a) .374; r = 8. (b) .4352; r = 6. (c) .2231 ; r = 4. (d) .2001 ; r = 3. 2. Express the decimal fraction .296 as a radix fraction for r = 5. 3. Express the decimal fraction .3405 as a radix fraction for r = Q. 34 4. Express — as a radix fraction for r = 4. ^ 128 5. Express as a radix fraction for r = 5. ^ 626 6. In what scale is 42 expressed as 1120? Solution : We seek r where r3 + r2 + 2 r = 42. This is equivalent to finding a positive integral root of the equation r3 + r2 + 2 r - 42 = 0. By synthetic division, 1 + 1 + 2 - 42(2 + 2 + 6 + 16 + 3 4- 8-26 1 + 1+ 2-4213 + 3 + 12 + 42 + 4+14 Thus 3 is the value sought. Check : 3^ + 32 + 2 • 3 = 27 + 9 + 6 = 42. 7. In what scale is 2704 denoted by 20304 ? 8. In what scale is 256 denoted by 10000 ? 9. In what scale is .1664 denoted by .0404 ? 10. Show that 1331 is a perfect cube. 235. Duodecimals. We may apply some of the foregoing processes to mensuration. If we take one foot as a unit and the radix as 12, we may express distances in the so-called duodecimal notation. Thus 2 ft. 6 in. is represented in the duodecimal scale by 2.6. Since in a scale of radix r we need ?• — 1 symbols, we will let 10 = ^ and 11 = e. Thus 21 ft. 10 in. would be expressed in duodecimal notation as 19.^. We may now find areas and vol- umes in this notation much more readily than by the usual method of converting all distances to inches. SCALES OF NOTATION 285 Example. Multiply 8 ft. 3 in. by 3 ft. 10 in. We multiply 8 • 3 by 3.i 8.3 in the duodecimal scale. To convert the result to square 3-t feet and square inches we must keep in mind that 27.76 6t6 ^ 2 . 12 + 7 + tV + xf^ = 31 sq. ft. 90 sq. in., since 144 — — square inches equal one square foot. This example suggests the following method of multiplying distances : EuLE. Express the distances in duodecimal notation with the foot as a unit. Multiply in the scale for which r = 12. In the product change the part on the left of the point from duodecimal to decimal scale. Multiply the digit following the point hy 12^ and add to the last figure to obtain the square inches in the result. EXERCISES 1. Multiply the following : (a) 13 ft. 4 in. by 67 ft. 11 in. Solution: 11.4 57.e 1028 794 568 636.68 = 905 sq. ft. 80 sq. in. (b) 10 ft. 6 in. by 12 ft. 2 in. (c) 8 ft. 4 in. by 11 ft. 11 in. (d) 23 ft. 6 in. by 47 ft. 8 in. (e) 41 ft. 6 in. by 36 ft. 1 in. 2. What is the area of a room 16 ft. 2 in. by 10 ft. 3 in. ? 3. What is the area of a walk 60 ft. 6 in. by 3 ft. 3 in. ? 4. What is the area of a city lot 62 ft. 6 in. by 163 ft. 7 in. ? ^4 UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. SEP 7 'S5h Ll. 100ct'55CT ,5^ov'56?W RECD LD NOV 1 ^9^^ «?Jul'57lVlH JUL 3 wa/ RECD LCi DEC 18 135f LD 21-100ot-7,'52 (A2528s16)476 AUG as* ^^;^« UNIVERSITY OF CALIFORNIA UBRARY ;':f:^